In this blog, I am posting two types of material.

One study guides on each chapter of the book H C Verma. I am giving the JEE syllabus relevant to the chapter, contents of the chapter, a brief on the concepts covered in the chapter and JEE questions from the chapter.

In the formula revision posts, the formulas given in each chapter are being collected for a quick reference and revision.

You can access both types of posts by clicking the chapter name in labels.

I am myself now reading first, then spending time on doing some problems and then do the posting. Currently I am putting a target of 31st May 2008 for completing all the chapters.

## Wednesday, February 20, 2008

## Friday, February 8, 2008

### IIT JEE Physics Formula Revision 3. Rest and motion :kinematics

1. v

where v

2. v = limΔt→0 Δs/Δt = ds/dt

where v = instantaneous speed

3. s = ∫vdt from t1 to t2

where s = distance travelled during time t1 to t2

4.

Where

t2, t1 = time instants

5.

6. v = |d

Where v = magnitude of velocity

s = displacement, for small time intervals magnitude of displacement will be equal to distance.

7.

8.

For Motion in straight line

9. Velocity is

10.

11. acceleration is

For constant acceleration

12. v = u+at

13. x = distance moved in time t = ut+½at²

14. v² = u²+2ax

For motion in plane

22. Time of flight of the projectile = (2u sin θ)/g

23. OB = (u²sin 2θ)/g

24. t = (u sin θ)/g

At t vertical component of velocity is zero.

25. Maximum height reached by the projectile (in t = (u sin θ)/g) = (u² sin²θ)/2g

27.

Where

we can rewrite above equation as

28.

_{av}= s/(t2 - t1)where v

_{av}= average speed2. v = limΔt→0 Δs/Δt = ds/dt

where v = instantaneous speed

3. s = ∫vdt from t1 to t2

where s = distance travelled during time t1 to t2

4.

**v**_{av}= (**r2**-**r1**)/(t2-t1)Where

**v**_{av}= average velocity**r2**,**r1**= position vectors of a particlet2, t1 = time instants

5.

**v**= lim Δt→0 Δ**r**/Δt = d**r**/dt6. v = |d

**r**|/dt = ds/dtWhere v = magnitude of velocity

s = displacement, for small time intervals magnitude of displacement will be equal to distance.

7.

**a**_{av}= (**v2**-**v1**)/(t2-t1)8.

**a**= lim Δt→0 Δ**v**/Δt = d**v**/dtFor Motion in straight line

9. Velocity is

**v**= dx/dt10.

**a**= dv/dt11. acceleration is

**a**= dv/dt = d²x/dt²For constant acceleration

12. v = u+at

13. x = distance moved in time t = ut+½at²

14. v² = u²+2ax

For motion in plane

22. Time of flight of the projectile = (2u sin θ)/g

23. OB = (u²sin 2θ)/g

24. t = (u sin θ)/g

At t vertical component of velocity is zero.

25. Maximum height reached by the projectile (in t = (u sin θ)/g) = (u² sin²θ)/2g

27.

**V**(B,S) =**V**(B,S')+**V**(S',S)Where

**V**(B,S) = velocity of body wrt to S)**V**(B,S') = velocity of body wrt to S')**V**(S',S) = velocity of S' wrt to S)we can rewrite above equation as

28.

**V**(B,S') =**V**(B,S)-**V**(S',S)### IIT JEE Physics Formula Revision 5. Newtons law of motion

1. Acceleration vectoS a(P,S) = Acceleration vector a(P,S') if S is an inertial frame and S' is a frame moving uniformly with respect to S.

### IIT JEE Physics Formula Revision 6. Friction

1. Normal force = N

Where

M = mass of the object

g = acceleration due to gravity

2. f

where

f

µ

3. f

where

f

µ

The actual static friction can be less than maximum static friction if the force applied is less than f

4. On an adjustable inclined plane, a block is kept and the angle is gradually increased so that the block begins to move.

Then f

N

There coefficient of static friction µ

µ

= tan θ = h/d

where θ = angle of incline when the block starts moving

m = mass of the block placed on incline

h = height of the incline

d= length of the incline

5. T find kinetic friction the angle of the incline is slightly reduced and the block is made to move with uniform velocity

In this case

µ

(h is going to decrease and d is going to increase)

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_{f}= MgWhere

M = mass of the object

g = acceleration due to gravity

2. f

_{k}= µ_{k}N_{f}where

f

_{k}= magnitude of kinetic frictionµ

_{k}= coefficient of kinetic friction3. f

_{max}= µ_{s}N_{f}where

f

_{max}= maximum static frictionµ

_{s}= coefficient of static frictionThe actual static friction can be less than maximum static friction if the force applied is less than f

_{max}.4. On an adjustable inclined plane, a block is kept and the angle is gradually increased so that the block begins to move.

Then f

_{max}= mg sin θN

_{f}= mg cos θThere coefficient of static friction µ

_{s}µ

_{s}= f_{max}/ N_{f}= tan θ = h/d

where θ = angle of incline when the block starts moving

m = mass of the block placed on incline

h = height of the incline

d= length of the incline

5. T find kinetic friction the angle of the incline is slightly reduced and the block is made to move with uniform velocity

In this case

µ

_{k}= coefficient of kinetic friction = tan θ’ = h’/d’(h is going to decrease and d is going to increase)

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### IIT JEE Physics Formula Revision 7. Circular motion

**Angular variables**

θ = angular position of the particle

ω = angular velocity = d θ/dt = lim∆t→0 ∆θ/∆t

α = angular acceleration = d ω/dt = d²θ/dt²

If the angular acceleration is constant, formulas similar in form to linear formulas can be used to find the angular variables:

θ = ω

_{0}t + ½ αt²

ω = ω

_{0}+ αt

ω² = ω

_{0}² + 2 α θ

**Linear variables of circular motion**

s = Linear distance traveled by the particle in circular motion

∆s = Linear distance traveled by the particle in circular motion in time ∆t

∆s = r∆θ

Where

r = radius of the circle over which the particle is moving

∆θ = angular displacement in time ∆t

∆s/∆t = r∆θ/∆t

v = r ω

where

v = linear speed of the particle

a

_{t}= rate of change of speed of the particle in circular motion

a

_{t}= dv/dt = rdω/dt = r α

**Unit Vectors along the Radius and Tangent on a point on the circle on which the particle is moving.**

If x-axis is horizontal and y axis is vertical (normal representation)

**e**= unit vector along the radius at a point on the circle

_{r}**e**=

_{r}**i**cos θ+

**j**sin θ

**e**= unit vector along the tangent at a point on the circle

_{t}**e**= -

_{t}**i**sin θ+

**j**cos θ

Position vector of the point P

If the angular position is θ, x coordinate is r cos θ, and y coordinate is r sin θ.

Hence position vector is

**r**= r(

**i**cos θ +

**j**sin θ)

Differentiating position vector with respect to time we get velocity

**v**= r ω(-

**i**sin θ +

**j**cos θ)

Differentiating v with respect to time

**a**= dv/dt = - ω²r

**e**+ dv/dt

_{r}**e**

_{t}**Uniform Circular Motion**

If the particle moves in the circle with a uniform speed (v = constant), it is uniform

circular motion.

In this case dv/dt = 0

Hence

**a**= dv/dt = - ω²r

**e**

_{r}Acceleration of the particle is in the direction of -

**e**, i.e, towards the centre of the circle. The magnitude of the acceleration is

_{r}a = ω²r = v²/r (as v = rω)

**Nonuniform circular motion**

In nonuniform circular motion, speed is not constant and the acceleration of the particle has both radial and tangential components.

**a**= - ω²r

**e**+ dv/dt

_{r}**e**

_{t}the radial component is a

_{r}= - ω²r = -v²/r

the tangential component is a

_{t}= dv/dt

The magnitude of the acceleration

a = √ (a

_{r}² + a

_{t}²)

= √[( ω²r) ² + (dv/dt) ²]

The direction of this acceleration makes an angle α with the radius connecting the centre of the circle with the point at that instant.

Tan α = (dv/dt)/ (ω²r)

**Forces in Circular Motion**

If a particle of mass m is moving along a circle with uniform speed, the force acting on it has to be

F = ma = mv²/r = m ω²r

This force is called centripetal force.

**Centrifugal Force**

It is equal to the centripetal force

F = mv²/r = m ω²r

**Circular Turnings on the Road**

When a turn is made, the speed has to be less so that friction can make the vehicle turn in stead of skid

For a safe turn we need to have

v²/r = f

_{s}/M

Or M v²/r ≤ f

_{s}

As f

_{s}≤ μ

_{s}Mg

M v²/r ≤ μ

_{s}Mg

That means v²/gr ≤ μ

_{s}

v≤√( μ

_{s}gr)

Instead of relying on friction, road are given an angle to provide centripetal force.

Tan θ = v²/gr

θ is the angle to be given to the road based v. Road are designed for assumed normal speed.

**Effect of Earth’s Rotation on Apparent Weight**

Axis of rotation of earth = The line joining the north and south pole = SN

Every point P on earth moves in circle.

Draw a perpendicular PC to SN from P.

Centre of the earth = O.

The distance between a point on the equator and centre of earth O = R radius of earth.

The angle between OP and PC is θ

θ is called the colatitude of the place where P is there.

If heavy particle of mass m is suspended through a string at P, the forces on it will be

1. gravitational attraction mg towards earth along PO

2. Centrifugal force = m ω² r

3. Tension in the string

The resultant of mg and m ω² r =

m√(g²+ ω

^{4}R²sin² θ-2g ω²R sin² θ) = mg’

The direction of this resultant force with OP is given by

Tan α = ω²Rsin θ cos θ/(g - ω²Rsin² θ)

A plumbline stays along g’

g’ = √(g²- ω²Rsin² θ(2g- ω²R ))

As 2g> ω²R; g’

At equator θ = 90° and

G’ = g- ω²R

At poles θ = 0° and hence

g’ = g

### IIT JEE Physics Formula Revision 8. Work and energy

1. K(v) = 1/2mv^2

2. dK = F.dr (both F and dr are vectors)

3. W = ∫F cosθ dr

2. dK = F.dr (both F and dr are vectors)

3. W = ∫F cosθ dr

### IIT JEE Physics Formula Revision 15. Wave motion and waves on a string

1. Equation of a wave travelling in the positive x-direction with a constant speed v.

The displacement of the particle at x at time t i.e., y(x,t) is generally abbreviated as y and the wave equation is written as

y = f(t - x/v) ... (1)

2. Equation of a wave travelling in the negative x-direction with a constant speed v.

The displacement of the particle at x at time t i.e., y(x,t) is generally abbreviated as y and the wave equation is written as

y = f(t + x/v) ... (2)

3. The wave equation in (1) can be written as y = f((vt-x)/v) which can be transformed into y = g(x-vt)...(3)

g is a different function. Function g can have the following meaning. If you put t = 0 in equation (3), you get the displacement of various particles at t = 0;

y(x, t = 0) = g(x)

If displacement at t = 0 of all particles of the string is represented by g(x) then the displacement of the particle at x at time t will be y = g(x-vt).

4. Similarly if the wave is travelling along the negative x-direction and the displacement of different particles at t = 0 is g(x), the displacement of the aprticle at x at time t will be

y = g(x+vt)

Function f in equation 1 and 2 represents the displacement of the point x = 0 as time passes,and g in (3) and (4) represents the diplacement at t = 0 of different particles.

5. sine wave or sinusoidal wave

When a person vibrates the left end of a string x = 0 in a simple harmonic motion, the equation of motion of this end may be written as

f(t) = A sin ωt ... (5)

A represents the amplitude

ω = the angular frequency

Time period of oscillation is T = 2π/ω

Frequency of oscillation = 1/T = ω/2π

This wave is called a sine wave or sinusoidal wave.

If the displacement of the particle at x = 0 is given by f(t) = A sin ωt, the displacement of the particle at x at time t will be given by

y = f(t - x/v) = A sin ω(t - x/v)...(6)

7. Velocity of the particle at x at time t is given by

ðy/ðt = Aω cos ω(t - x/v)...(7)

This velocity is different from the velocity of the wave. The wave moves on the string at a constant velocity v along the x axis, but the particles at various points move up and down with velocity ðy/ðt which changes with x and t according to equation (7).

8. λ = (v/ω)2π = vT .... (8)

9. v = λ/T = νλ ... (9)

where ν = 1/T is the frequency of the wave.

Alternative sine wave equations

y = A sin ω(t - x/v)

y = A sin (ωt - kx)... (10)

y = A sin 2π(t/T - x/λ)... (11)

y = A sin k(vt-x) ... (12)

General equation will be

y = A sin [ω(t - x/v)+ Φ)... (13)

Φ is the phase constant.

This equation will allow us to write equation based on the displacement of the left at t = 0.

The constant Φ will be π/2 is we choose t = 0at an instant when the left end reaches its extreme position y = A. The equation will then be

y = A cos ω(t - x/v)... (14)

If t = 0 is taken at the instant when the left end is crossing the mean position from upward to downward direction, Φ will be π and the equation will be

y = A sin (kx-ωt)... (15)

16. Velocity of a wave on a string

v = √(F/μ)

F = tension in the string

μ = linear mass density of the string ... (16)

17. Power transmitted along the string by a sine wave

Pav = 1/2 [ω² A² F²/v]... (17)

The displacement of the particle at x at time t i.e., y(x,t) is generally abbreviated as y and the wave equation is written as

y = f(t - x/v) ... (1)

2. Equation of a wave travelling in the negative x-direction with a constant speed v.

The displacement of the particle at x at time t i.e., y(x,t) is generally abbreviated as y and the wave equation is written as

y = f(t + x/v) ... (2)

3. The wave equation in (1) can be written as y = f((vt-x)/v) which can be transformed into y = g(x-vt)...(3)

g is a different function. Function g can have the following meaning. If you put t = 0 in equation (3), you get the displacement of various particles at t = 0;

y(x, t = 0) = g(x)

If displacement at t = 0 of all particles of the string is represented by g(x) then the displacement of the particle at x at time t will be y = g(x-vt).

4. Similarly if the wave is travelling along the negative x-direction and the displacement of different particles at t = 0 is g(x), the displacement of the aprticle at x at time t will be

y = g(x+vt)

Function f in equation 1 and 2 represents the displacement of the point x = 0 as time passes,and g in (3) and (4) represents the diplacement at t = 0 of different particles.

5. sine wave or sinusoidal wave

When a person vibrates the left end of a string x = 0 in a simple harmonic motion, the equation of motion of this end may be written as

f(t) = A sin ωt ... (5)

A represents the amplitude

ω = the angular frequency

Time period of oscillation is T = 2π/ω

Frequency of oscillation = 1/T = ω/2π

This wave is called a sine wave or sinusoidal wave.

If the displacement of the particle at x = 0 is given by f(t) = A sin ωt, the displacement of the particle at x at time t will be given by

y = f(t - x/v) = A sin ω(t - x/v)...(6)

7. Velocity of the particle at x at time t is given by

ðy/ðt = Aω cos ω(t - x/v)...(7)

This velocity is different from the velocity of the wave. The wave moves on the string at a constant velocity v along the x axis, but the particles at various points move up and down with velocity ðy/ðt which changes with x and t according to equation (7).

8. λ = (v/ω)2π = vT .... (8)

9. v = λ/T = νλ ... (9)

where ν = 1/T is the frequency of the wave.

Alternative sine wave equations

y = A sin ω(t - x/v)

y = A sin (ωt - kx)... (10)

y = A sin 2π(t/T - x/λ)... (11)

y = A sin k(vt-x) ... (12)

General equation will be

y = A sin [ω(t - x/v)+ Φ)... (13)

Φ is the phase constant.

This equation will allow us to write equation based on the displacement of the left at t = 0.

The constant Φ will be π/2 is we choose t = 0at an instant when the left end reaches its extreme position y = A. The equation will then be

y = A cos ω(t - x/v)... (14)

If t = 0 is taken at the instant when the left end is crossing the mean position from upward to downward direction, Φ will be π and the equation will be

y = A sin (kx-ωt)... (15)

16. Velocity of a wave on a string

v = √(F/μ)

F = tension in the string

μ = linear mass density of the string ... (16)

17. Power transmitted along the string by a sine wave

Pav = 1/2 [ω² A² F²/v]... (17)

### IIT JEE Physics Formula Revision 23. Heat and Temperature

1. conversion formula is F = 32 + 9C/5

2. The temperature of triple point of water is assigned a value of 273.16 K

The temperature of ice point on the ideal gas scale is 273.15 K and of the steam point is T = 373.15 K. The interval between the two is 100 K.

Centigrade scale or Celsius scale is defined to have ice point at 0°C and steam point at 100°C. The interval is 100°C.

Hence if θ represents the Celsius or centigrade temperature

V = T – 273.15 K

3. Ideal gas equation

pV = nRT

n = the amount of gas in moles

R = universal constant = 8.314 J/mol-K

Thermal expansion

Average coefficient of linear expansion

Average (α) = (1/L)*∆L/∆T

Coefficient of linear expansion at temperature T

α = Lim (∆T->0)(1/L)*∆L/∆T = (1/L)dL/dT

Suppose the length of a rod is L0 at 0° C and Lθ at temperature θ measured in Celsius. If α is small and constant over the given temperature interval,

α = Lθ-L0/L0*θ or

Lθ = L0(1+αθ)

The coefficient of volume expansion is defined in a similar way.

γ = (1/V)dV/dT

It is also known as coefficient of cubical expansion.

Vθ = V0(1+γθ)

γ = 3α

2. The temperature of triple point of water is assigned a value of 273.16 K

The temperature of ice point on the ideal gas scale is 273.15 K and of the steam point is T = 373.15 K. The interval between the two is 100 K.

Centigrade scale or Celsius scale is defined to have ice point at 0°C and steam point at 100°C. The interval is 100°C.

Hence if θ represents the Celsius or centigrade temperature

V = T – 273.15 K

3. Ideal gas equation

pV = nRT

n = the amount of gas in moles

R = universal constant = 8.314 J/mol-K

Thermal expansion

Average coefficient of linear expansion

Average (α) = (1/L)*∆L/∆T

Coefficient of linear expansion at temperature T

α = Lim (∆T->0)(1/L)*∆L/∆T = (1/L)dL/dT

Suppose the length of a rod is L0 at 0° C and Lθ at temperature θ measured in Celsius. If α is small and constant over the given temperature interval,

α = Lθ-L0/L0*θ or

Lθ = L0(1+αθ)

The coefficient of volume expansion is defined in a similar way.

γ = (1/V)dV/dT

It is also known as coefficient of cubical expansion.

Vθ = V0(1+γθ)

γ = 3α

### IIT JEE Physics Formula Revision 24. Kinetic theory of Gases

1. p = 1/3*ρ*Mean(v²)

p = pressure of gas

ρ = density of gas

Mean(v²) = mean square speed of molecules of gas

2. pV = 1/3*M*Mean(v²)

V = Volume of gas

M = Total mass of the gas taken

3. pV = 1/3*Nm*Mean(v²)

N = total number of molecules in the sample

m = mass of one molecule

4. Mean(v²) = ΣV²/N

Square root of Mean(v²) is also called root-mean square speed v

Equation 1 can be written as p = 1/3*ρ*v²

So v

6. In thermal equilibrium in a mixture of two gases, the average kinetic energies of all molecules are equal. If v1 an v2 are rms speeds of the molecules of A and B

½m1v1² = ½m2v2²

7. pV = Nkt

k is known as the Boltzmann constant.

Its value is 1.38*10^(-23) J/K

R = NA*k (NA is Avogadro's number = 6.02*10^23

R = 8.314 J/mol-K

8. pV = nRT

9. RMS spped in terms of temperature

v(rms) = SQRT(3kT/m)

10. v(rms) = SQRT(3RT/M0)

11. Average kinetic energy of a molecule

½mv(rms)² = (3/2) kT

12. Total kinetic energy of all molecules

U = (3/2)nNA kT

or

U = 3/2nRT ( R = NA*k)

n = number of moles of substance

13. maxwells speed distribution law

dN = 4 π[m/2 πkT]

14. The speed at which dN/dv is maximum is called the most probable speed.

v

15. van der Waals equation for real gases

For an ideal gas

pV = nRT

This is called the equation of state for that substance

For a real gas

[p + a/V²][V-b] = nRT

where a and b are small positive constants.

a is related to the average force of all attracion between the molecules.

bis related to the total volume of the molecules.

This equation is given by van der Waals.

16. Relative humidity = RH =

Amount of water vapour present in a given volume of air at a given temperature

--------------------------------------------------------------------------------------------------

Amount of water vapour required to saturate the same volume of air at the same tempeture

RH may also be defined as

Vapour pressure of air/SVP at the same temperature

SVP = Saturation vapour pressure: The pressure exerted by a saturated vapour is called saturation vapour pressure.

The RH may also be defined as

SVP at the dew point/SVP at the air temperature

As the vapour pressure of air at the actual temperature is equal to the SVP at the dew point.

p = pressure of gas

ρ = density of gas

Mean(v²) = mean square speed of molecules of gas

2. pV = 1/3*M*Mean(v²)

V = Volume of gas

M = Total mass of the gas taken

3. pV = 1/3*Nm*Mean(v²)

N = total number of molecules in the sample

m = mass of one molecule

4. Mean(v²) = ΣV²/N

Square root of Mean(v²) is also called root-mean square speed v

_{rms}Equation 1 can be written as p = 1/3*ρ*v²

_{rms}So v

_{rms}= SQRT(3p/ρ)6. In thermal equilibrium in a mixture of two gases, the average kinetic energies of all molecules are equal. If v1 an v2 are rms speeds of the molecules of A and B

½m1v1² = ½m2v2²

7. pV = Nkt

k is known as the Boltzmann constant.

Its value is 1.38*10^(-23) J/K

R = NA*k (NA is Avogadro's number = 6.02*10^23

R = 8.314 J/mol-K

8. pV = nRT

9. RMS spped in terms of temperature

v(rms) = SQRT(3kT/m)

10. v(rms) = SQRT(3RT/M0)

11. Average kinetic energy of a molecule

½mv(rms)² = (3/2) kT

12. Total kinetic energy of all molecules

U = (3/2)nNA kT

or

U = 3/2nRT ( R = NA*k)

n = number of moles of substance

13. maxwells speed distribution law

dN = 4 π[m/2 πkT]

^{3/2}v²e^{-(mv²/2kT)}dv14. The speed at which dN/dv is maximum is called the most probable speed.

v

_{p}= SQRT(2kT/m)15. van der Waals equation for real gases

For an ideal gas

pV = nRT

This is called the equation of state for that substance

For a real gas

[p + a/V²][V-b] = nRT

where a and b are small positive constants.

a is related to the average force of all attracion between the molecules.

bis related to the total volume of the molecules.

This equation is given by van der Waals.

16. Relative humidity = RH =

Amount of water vapour present in a given volume of air at a given temperature

--------------------------------------------------------------------------------------------------

Amount of water vapour required to saturate the same volume of air at the same tempeture

RH may also be defined as

Vapour pressure of air/SVP at the same temperature

SVP = Saturation vapour pressure: The pressure exerted by a saturated vapour is called saturation vapour pressure.

The RH may also be defined as

SVP at the dew point/SVP at the air temperature

As the vapour pressure of air at the actual temperature is equal to the SVP at the dew point.

### IIT JEE Physics Formula Revision 25. Calorimetry

1. Q = mS∆θ

S = specific heat capacity of a substance

2. Q = mC∆θ

C = molar specific heat capacity of a substance

3. W = JH

W = work measured in joule

H = Heat measured in calories

J = mechanical equivalent to heat

= joule/calorie = 4.186 J/cal

S = specific heat capacity of a substance

2. Q = mC∆θ

C = molar specific heat capacity of a substance

3. W = JH

W = work measured in joule

H = Heat measured in calories

J = mechanical equivalent to heat

= joule/calorie = 4.186 J/cal

### IIT JEE Physics Formula Revision 26. Laws of Thermodynamics

First law of thermodynamics and its applications (only for ideal gases) 26.1;

26.1 ΔU = ΔQ - ΔW or

ΔQ = ΔU + ΔW

26.2 ΔW =∫pΔV from V1 to V2

26.3 Work done in an Isothermal process:

W is given by the expression nRTln[V2/V1]

Work doen in an Isobaric process - pressure is constant

W = p[V2 - V1]

Work done in an isochoric process: volume is constant. No work is done by the system.

4. Efficiency of an engine

η = W/Q1 = 1 - Q2/Q1

Q1 = Amout of heat taken by the engine

Q2 = Heat rejected by the engine to low temperature bodies

5. Entropy

∆S = ∆Q/T

6. Sf - Si = ∫∆Q/T from i to f

26.1 ΔU = ΔQ - ΔW or

ΔQ = ΔU + ΔW

26.2 ΔW =∫pΔV from V1 to V2

26.3 Work done in an Isothermal process:

W is given by the expression nRTln[V2/V1]

Work doen in an Isobaric process - pressure is constant

W = p[V2 - V1]

Work done in an isochoric process: volume is constant. No work is done by the system.

4. Efficiency of an engine

η = W/Q1 = 1 - Q2/Q1

Q1 = Amout of heat taken by the engine

Q2 = Heat rejected by the engine to low temperature bodies

5. Entropy

∆S = ∆Q/T

6. Sf - Si = ∫∆Q/T from i to f

### IIT JEE Physics Formula Revision 27. Specific Heat of Capacities of Gases

1. Specific heat capacity of a substance s

s = ΔQ/mΔT

2. Specific heat capacity of a substance at constant volume is denoted by symbol sv

sv = [ΔQ/mΔT] heat given at constant volume

3. Specific heat capacity of a substance at constant pressure is denoted by symbol sp

sp = [ΔQ/mΔT] heat given at constant pressure

4. Molar heat capacity of a substance at constant volume is denoted by symbol Cv

Cv = [ΔQ/nΔT] heat given at constant volume

5. Molar heat capacity of a substance at constant pressure is denoted by symbol Cp

Cp = [ΔQ/mΔT] heat given at constant pressure

6. Relation between Cp and Cv for an ideal gas

Cp - Cv = R

or

Cp = Cv + R

7. Cv = i/n (dU/dT)

dU = n(Cv)dT

8. Taking energy to be zero at t = 0

U = n(Cv)T

9. pV

γ = Cp/Cv

10. In a reversible adiabatic process

T

11. In a reversible adiabatic process

TV

12. Work done in an adiabatic process

W = (p1V1 - p2V2)/(γ-1)

For monoatomic gas

U = n (3/2) RT

Cv = 3/2R

Cp = 5/2R

γ = 5/3 = 1.67

For diatomic gas without vibration

Cv = 5/2R

Cp = 7/2R

γ = 7/5 = 1.40

If the molecules vibrate

Cv = 7/2R

Cp = 9/2R

γ = 9/7 = 1.29

s = ΔQ/mΔT

2. Specific heat capacity of a substance at constant volume is denoted by symbol sv

sv = [ΔQ/mΔT] heat given at constant volume

3. Specific heat capacity of a substance at constant pressure is denoted by symbol sp

sp = [ΔQ/mΔT] heat given at constant pressure

4. Molar heat capacity of a substance at constant volume is denoted by symbol Cv

Cv = [ΔQ/nΔT] heat given at constant volume

5. Molar heat capacity of a substance at constant pressure is denoted by symbol Cp

Cp = [ΔQ/mΔT] heat given at constant pressure

6. Relation between Cp and Cv for an ideal gas

Cp - Cv = R

or

Cp = Cv + R

7. Cv = i/n (dU/dT)

dU = n(Cv)dT

8. Taking energy to be zero at t = 0

U = n(Cv)T

9. pV

^{γ}= constantγ = Cp/Cv

10. In a reversible adiabatic process

T

^{γ}/p^{γ-1}= constant11. In a reversible adiabatic process

TV

^{γ-1}= constant12. Work done in an adiabatic process

W = (p1V1 - p2V2)/(γ-1)

For monoatomic gas

U = n (3/2) RT

Cv = 3/2R

Cp = 5/2R

γ = 5/3 = 1.67

For diatomic gas without vibration

Cv = 5/2R

Cp = 7/2R

γ = 7/5 = 1.40

If the molecules vibrate

Cv = 7/2R

Cp = 9/2R

γ = 9/7 = 1.29

### IIT JEE Physics Formula Revision 28. Heat transfer

1. ∆Q/∆t = KA(T1-T2)/x

2. ∆Q/∆t = -KAdT/dx

3. x/KA is called the thermal resistance R.

∆Q/∆t is termed as heat current i

i = T1-T2/R

4. In the experimental determination of thermal conductivity of a solid

K = [xms(θ3-θ4)]/[A(θ1-θ2)t]

5. Wien's displacement law

λ

6.

The energy of thermal radiation emitted by per unit time by a black body of surface area A is given by

U = σAT

Where

σ = Stefan Boltzmann constant = 5.67*10

The energy of thermal radiation emitted by per unit time by a body (not black) of surface area A is given by

U = eσAT

e is a constant for the given surface called emissivity of the surface. Its value is between 0 and 1. It is one for black body and 0 for completely reflecting surface.

7. Net loss of thermal energy of body with temperature T kept in a room at temperature T

u = eσA(T

8.

dT/dt = -bA(T-T

b = a constant depends on the nature of the surface involved

a = surface area exposed of the body

T- T

2. ∆Q/∆t = -KAdT/dx

3. x/KA is called the thermal resistance R.

∆Q/∆t is termed as heat current i

i = T1-T2/R

4. In the experimental determination of thermal conductivity of a solid

K = [xms(θ3-θ4)]/[A(θ1-θ2)t]

5. Wien's displacement law

λ

_{m}T = b = constant6.

**Stefan-Boltzmann Law**The energy of thermal radiation emitted by per unit time by a black body of surface area A is given by

U = σAT

^{4}Where

σ = Stefan Boltzmann constant = 5.67*10

^{-8}W/m²-K^{4}The energy of thermal radiation emitted by per unit time by a body (not black) of surface area A is given by

U = eσAT

^{4}e is a constant for the given surface called emissivity of the surface. Its value is between 0 and 1. It is one for black body and 0 for completely reflecting surface.

7. Net loss of thermal energy of body with temperature T kept in a room at temperature T

_{0}u = eσA(T

^{4}- T_{0}^{4})8.

**Newton’s law of cooling**dT/dt = -bA(T-T

_{0})b = a constant depends on the nature of the surface involved

a = surface area exposed of the body

T- T

_{0}) = temperature difference between the body and surrounding### IIT JEE Physics Formula Revision 29. Electric Field and Potential

For revision points of the chapter

http://iit-jee-physics.blogspot.com/2008/03/concept-review-ch-29-electric-field-and.html

The charge on a proton is

e = 1.60218*10^-19 C

Coulomb's formula for the electric force between two charges.

(1) F = k*q1*q2/r²

q1,q2 charges

r = separation between charges

k = constant

In SI units k is measured to be 8.98755*10^9 N-m²/C²

The constant k is often written as 1/4πε

The constant ε

ε

The intensity of field is defined as Vector E = Vector F/q

(3)

Electric field due to a point charge (Q)

(4) E = F/q = Q/4πε

(5) U(r2) - U(r1) = -W = [q1*q2/4πε

Choosing potential energy at infinite separation as zero

(6) U(r) = U(r) - U(∞) = [q1*q2/4πε

= q1*q2/4πε

(7) VB - VA = (UB - UA)/q

We can define absolute electric potential at any point by choosing a reference point P and saying that the potential at this point is zero.

(8) VA = VA - VP = (UA - UP)/q

The potential due to a point charge Q (placed at A) at a point P with distance AP = r is

(9) VP = Q/4πε

The electric potential at a point due to more than one charge in the system is obtained by finding potential due to individual charges and then adding them.

10. Relation between electric field and potential

dV = -

Integrating between limits r1 and r2

V2 – V1 = - ∫

If we choose r1 as infinity (reference point)

V(r) = - ∫

Hence we can calculate potential V if we know E and r.

If we know V we can find E through the relation

Ex = - ∂V/∂x

Ey = -∂V/∂y

Ez = -∂V/∂z

We can find the x,y and z components of E.

E can be written as Ex

We can write dV as dV = -E dr cos θ where θ is the angle between the field E and the small displacement dr.

-dV/dr = E cos θ

15. Electric Dipole

Electric dipole moment

It is defined as a vector

where distance vector is the vector joining the negative charge to the positive charge.

The line along the direction of the dipole moment is called the axis of the dipole.

16. Electric potential due to a dipole at a point P

Point is at distance r from the centre of the diploe (d/2) and theline joining the point P to the centre of the dipole make an angle θ with the direction of dipole movement (from –q to +q)

Potential at P due to charge –q = - [1/4πε

Potential at P due to charge q = [1/4πε

Net potential due to q and –q = [1/4πε

17. Electric field due to a dipole

E

E

Resultant electric field at P = E= √ (E

= [1/4πε

19. Torque on an electric dipole placed in an electric field.

If the dipole axis makes an angle θ with the electric field magnitude of the torque = | Γ| = pE sin θ

In vector notation

20. Potential energy of a dipole placed in a uniform electric field

dipole axis makes an angle θ with the electric field magnitude of the torque

Change in potential energy = U(θ) – U(90°) = -pE cos θ = -

If we choose the potential energy of the dipole to be zero when θ = 90° , above equation becomes

U(θ) = -pE cos θ = -

http://iit-jee-physics.blogspot.com/2008/03/concept-review-ch-29-electric-field-and.html

The charge on a proton is

e = 1.60218*10^-19 C

Coulomb's formula for the electric force between two charges.

(1) F = k*q1*q2/r²

q1,q2 charges

r = separation between charges

k = constant

In SI units k is measured to be 8.98755*10^9 N-m²/C²

The constant k is often written as 1/4πε

_{0}.The constant ε

_{0}is called the permittivity of the space and its value isε

_{0}= 8.85419*10^-12 C²/N-m²The intensity of field is defined as Vector E = Vector F/q

(3)

**E**=**F**/qElectric field due to a point charge (Q)

(4) E = F/q = Q/4πε

_{0}r²(5) U(r2) - U(r1) = -W = [q1*q2/4πε

_{0}]* [(1/r2) - (1/r1)]Choosing potential energy at infinite separation as zero

(6) U(r) = U(r) - U(∞) = [q1*q2/4πε

_{0}]* (1/r)= q1*q2/4πε

_{0}r(7) VB - VA = (UB - UA)/q

We can define absolute electric potential at any point by choosing a reference point P and saying that the potential at this point is zero.

(8) VA = VA - VP = (UA - UP)/q

The potential due to a point charge Q (placed at A) at a point P with distance AP = r is

(9) VP = Q/4πε

_{0}rThe electric potential at a point due to more than one charge in the system is obtained by finding potential due to individual charges and then adding them.

10. Relation between electric field and potential

dV = -

**E**.**dr**Integrating between limits r1 and r2

V2 – V1 = - ∫

**E**.**dr**If we choose r1 as infinity (reference point)

V(r) = - ∫

**E**.**dr**(from infinity to r)Hence we can calculate potential V if we know E and r.

If we know V we can find E through the relation

Ex = - ∂V/∂x

Ey = -∂V/∂y

Ez = -∂V/∂z

We can find the x,y and z components of E.

E can be written as Ex

**i**+Ey**j**+ Ez**k**We can write dV as dV = -E dr cos θ where θ is the angle between the field E and the small displacement dr.

-dV/dr = E cos θ

15. Electric Dipole

Electric dipole moment

It is defined as a vector

**p**= q*distance vector**d**where distance vector is the vector joining the negative charge to the positive charge.

The line along the direction of the dipole moment is called the axis of the dipole.

16. Electric potential due to a dipole at a point P

Point is at distance r from the centre of the diploe (d/2) and theline joining the point P to the centre of the dipole make an angle θ with the direction of dipole movement (from –q to +q)

Potential at P due to charge –q = - [1/4πε

_{0}][q/(r + (dcos θ)/2)]Potential at P due to charge q = [1/4πε

_{0}][q/(r - (dcos θ)/2)]Net potential due to q and –q = [1/4πε

_{0}](qd cos θ)/r²17. Electric field due to a dipole

E

_{r}= [1/4πε_{0}](2p cos θ)/r³E

_{θ}= [1/4πε_{0}](p sin θ)/r³Resultant electric field at P = E= √ (E

_{r}²+ E_{θ}²)= [1/4πε

_{0}](p/r³)√(3 cos²θ + 1)19. Torque on an electric dipole placed in an electric field.

If the dipole axis makes an angle θ with the electric field magnitude of the torque = | Γ| = pE sin θ

In vector notation

**Γ**=**p**×**E**20. Potential energy of a dipole placed in a uniform electric field

dipole axis makes an angle θ with the electric field magnitude of the torque

Change in potential energy = U(θ) – U(90°) = -pE cos θ = -

**p**.**E**If we choose the potential energy of the dipole to be zero when θ = 90° , above equation becomes

U(θ) = -pE cos θ = -

**p**.**E**### IIT JEE Physics Formula Revision 30. Gauss's Law

1. ∆Φ = E ∆s cos θ

∆Φ is the flux of the electric field (E) through the surface ∆s. θ is the angle between positive normal to the surface and electric field.

Using the techniques of integration flux over a surface is:

Φ = ∫

Flux over a closed surface

Φ = ∮

2. Solid angle

Ω = S/r²

Ω = solid angle (dimensionless figure)

S = the area of the part of sphere intercepted by the cone

r = radius of the sphere assumed on which we are assuming the cone

A complete circle subtends an angle 2 π

Any closed surface subtends a solid angle 4 π at the centre.

How much is the angle subtended by a closed plane curve at an external point? Zero.

3. Gauss's law

The flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by ε

In symbols

∮

q

Where

q

ε

4. Electric field due to a uniformly charged sphere

A total charge Q is uniformly distributed in a spherical volume of radius R. what is the electric field at a distance r from the centre of the charge distribution outside the sphere?

E = Q/4π ε

5. Field at an internal point

At centre E = O

At any other point inside (rless than R) radius of the sphere

E = Qr/4π ε

6. c. Electric field due to a linear charge distribution

The linear charge density (charge per unit length) is λ.

Electric field at a distance r from the linear charge distribution

E = λ/2π ε

7. d. Electric field due to a plane sheet of charge

Plane sheet with charge density (charge per unit area) σ.

Field at distance d from the sheet = E = σ/2ε

We see that the field is uniform and does not depend on the distance from the charge sheet. This is true as long as the sheet is large as compared to its distance from P.

8. Electric field due to a charged conducting surface

To find the field at a point near this surface but outside the surface having charge density σ.

E = σ/ ε

Updated 12 October 2008

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∆Φ is the flux of the electric field (E) through the surface ∆s. θ is the angle between positive normal to the surface and electric field.

Using the techniques of integration flux over a surface is:

Φ = ∫

**E**.**∆s**Flux over a closed surface

Φ = ∮

**E**.**∆s**(* Intregration over a closed surface)2. Solid angle

Ω = S/r²

Ω = solid angle (dimensionless figure)

S = the area of the part of sphere intercepted by the cone

r = radius of the sphere assumed on which we are assuming the cone

A complete circle subtends an angle 2 π

Any closed surface subtends a solid angle 4 π at the centre.

How much is the angle subtended by a closed plane curve at an external point? Zero.

3. Gauss's law

The flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by ε

_{0}.In symbols

∮

**E**.**∆s**(* Intregration over a closed surface) =q

_{in}/ε_{0}.Where

q

_{in}= charge enclosed by the closed surfaceε

_{0}= emittivity of the free space4. Electric field due to a uniformly charged sphere

A total charge Q is uniformly distributed in a spherical volume of radius R. what is the electric field at a distance r from the centre of the charge distribution outside the sphere?

E = Q/4π ε

_{0}r²5. Field at an internal point

At centre E = O

At any other point inside (rless than R) radius of the sphere

E = Qr/4π ε

_{0}R³6. c. Electric field due to a linear charge distribution

The linear charge density (charge per unit length) is λ.

Electric field at a distance r from the linear charge distribution

E = λ/2π ε

_{0}r7. d. Electric field due to a plane sheet of charge

Plane sheet with charge density (charge per unit area) σ.

Field at distance d from the sheet = E = σ/2ε

_{0}We see that the field is uniform and does not depend on the distance from the charge sheet. This is true as long as the sheet is large as compared to its distance from P.

8. Electric field due to a charged conducting surface

To find the field at a point near this surface but outside the surface having charge density σ.

E = σ/ ε

_{0}Updated 12 October 2008

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### IIT JEE Physics Formula Revision - Ch. 31. Capacitors

1. For a given capacitor, the charge Q on the capacitor is proportional to the potential difference V between the two plates

So Q α V

or Q = CV

C is called the capacitance of the capacitor.

SI unit of capacitance is coulomb/volt which is written as farad. The symbol F is used for it.

C = ε

A = area of the flat plates (each used in the capacitor)

d = distance between the plate

It consists of a solid or hollow spherical conductor surrounded by another concentric hollow spherical conductor.

If inner sphere radius is R1 and Outer sphere radius is R2

Inner sphere is given positive charge and outer sphere negative charge.

C = 4πε

If the capacitor is an isolated sphere (outer sphere is assumed to be at infinity, hence R

C = 4πε

V becomes Q/C = Q/4πε

V = potential

Parallel limit: if both R1 and R2 are made large but R2-R1 = d is kept fixed

we can write

4πR1R2 = 4πR² = A; where R is approximately the radius of each sphere, and A is the surface area of the sphere.

C = ε

If inner cylinder radius is R1 and Outer cylinder radius is R2 and length is l,

Inner cylinder is given positive charge and outer cylinder negative charge

C = 2πε

Series combination

1/C = 1/C1 + 1/C2 + 1/C3 ...

Parallel combination

C = C1 + C2 + C3

6.

Plates on a parallel capacitor attract each other with a force

F = Q²/2Aε

7.

Capacitor of capacitance C has a stored energy

U = Q²/2C = CV²/2 = QV/2

Where Q is the charge given to it.

8. Change in capacitance of a capacitor with dielectric in it.

Polarization P (which is dipole moment induced per unit volume - where is the dipole? in the diectric slab as the two sides have opposite charges)

If σ

The dipole moment (q*Vr(d)) of the slab is then charge*l (distance between faces)

= σ

where

A is area of cross section of the dielectric slab

As polarization is defined as dipole moment induced per unit volume,

P = σ

= σ

The induced surface charge density is equal in magnitude to the polarization P.

9.

C = KC

where C

Because K>1, the capacitance of a capacitor is increased by a factor of K when the space between the parallel plates is filled with a dielectric.

Q

Q

Q = Applied charge

K = dielectric constant

10.

∮K

Where integration is over the surface, E and dS are vectors, Q

The law can also be written as

∮

where D = Eε

E = electric field and P is polarisation

11. Electric field due to a point charge placed inside a dielectric

E = q/4πε

Energy in the electric field in a dielectric

u = ½Kε

So Q α V

or Q = CV

C is called the capacitance of the capacitor.

SI unit of capacitance is coulomb/volt which is written as farad. The symbol F is used for it.

**2. For parallel plate capacitor**C = ε

_{0}A/dA = area of the flat plates (each used in the capacitor)

d = distance between the plate

**3. Spherical capacitor**It consists of a solid or hollow spherical conductor surrounded by another concentric hollow spherical conductor.

If inner sphere radius is R1 and Outer sphere radius is R2

Inner sphere is given positive charge and outer sphere negative charge.

C = 4πε

_{0}R_{1}R_{2}/[R_{2}-R_{1}]If the capacitor is an isolated sphere (outer sphere is assumed to be at infinity, hence R

_{2}is infinity andC = 4πε

_{0}R_{1}V becomes Q/C = Q/4πε

_{0}R_{1}V = potential

Parallel limit: if both R1 and R2 are made large but R2-R1 = d is kept fixed

we can write

4πR1R2 = 4πR² = A; where R is approximately the radius of each sphere, and A is the surface area of the sphere.

C = ε

_{0}A/d; where A = 4πR1R2 = 4πR²**4. Cylindrical Capacitor**If inner cylinder radius is R1 and Outer cylinder radius is R2 and length is l,

Inner cylinder is given positive charge and outer cylinder negative charge

C = 2πε

_{0}l/ln(R_{2}/R_{1})**5. Combination of capacitors**Series combination

1/C = 1/C1 + 1/C2 + 1/C3 ...

Parallel combination

C = C1 + C2 + C3

6.

**Force between plates of a capacitor**Plates on a parallel capacitor attract each other with a force

F = Q²/2Aε

_{0}7.

**Energy stored in a capacitor**Capacitor of capacitance C has a stored energy

U = Q²/2C = CV²/2 = QV/2

Where Q is the charge given to it.

8. Change in capacitance of a capacitor with dielectric in it.

Polarization P (which is dipole moment induced per unit volume - where is the dipole? in the diectric slab as the two sides have opposite charges)

If σ

_{p}is the magnitude of the induced charge per unit area on the faces.The dipole moment (q*Vr(d)) of the slab is then charge*l (distance between faces)

= σ

_{p}Al.where

A is area of cross section of the dielectric slab

As polarization is defined as dipole moment induced per unit volume,

P = σ

_{p}Al/Al (Al = volume of slab)= σ

_{p}The induced surface charge density is equal in magnitude to the polarization P.

9.

**Capacitance of a parallel plate capacitor with dielectric**C = KC

_{0}where C

_{0}is capacitance of a similar capacitor without dielectric.Because K>1, the capacitance of a capacitor is increased by a factor of K when the space between the parallel plates is filled with a dielectric.

**Magnitude of induced charge in term of K**Q

_{P}= Q[1 - (1/K)]Q

_{P}= induced charge in the dielectricQ = Applied charge

K = dielectric constant

10.

**Gauss's law when dielectric materials are involved**∮K

**E**.**dS**= Q_{free}/ε_{0}Where integration is over the surface, E and dS are vectors, Q

_{free}is the free charge given (charge due to polarisation is not considered) and K is dielectric constant.The law can also be written as

∮

**D**.**dS**= Q(free)where D = Eε

_{0}+ P; E and P are vectorsE = electric field and P is polarisation

11. Electric field due to a point charge placed inside a dielectric

E = q/4πε

_{0}Kr²Energy in the electric field in a dielectric

u = ½Kε

_{0}E²### IIT JEE Physics Formula Revision 32. Electric Current in Conductors

1. Average current i(bar) = ∆Q/∆t

The current at time t = I = lim ∆t→0 ( ∆Q/∆t) = dQ/dt

Q is charge, t is time

Average current density j (bar) = Δi/ΔA

i is current and A is area of the conductor

The current density at a point P is

j = lim ∆t→0 (Δi/ΔA) = di/dS

If current i is uniformly distributed over an area S and is perpendicular to it

j = i/S

For a finite area i = ∫j.dS

Where

j = density of current (vector)

dS = area (vector)

3. If τ be the average time between successive collisions, the distance drifted during this period is

l = ½ a(τ) = ½ (eE/m)( τ) ²

The drift speed is

v

τ the average time between successive collisions, is constant for a given material at a given temperature.

Relation between current density and drift speed

j = i/A = nev

j = σE

E is field and σ is electrical conductivity of the material.

Resistivity of a material ρ = 1/σ

V = voltage difference between the ends of a conductor = El (l = length of the conductor)

V = Ri

R = resistance of the conductor = ρ*l/A

1/R is called conductance

5. As temperature of a resistor increases its resistance increases. The relation can be expressed as

R(T) = R(T

α is called temperature coefficient of resistivity.

6. The work done by the battery force per unit charge is

Є = W/q = F

This Є is called the emf of the battery. Please note that emf is work done/charge.

If nothing is externally connected

F

F

V = potential difference between the terminals

As Є = W/q = F

Therefore Є = V

7. Thermal energy produced in a register

U = i²Rt

Power developed = P = U/t = i²R = Vi

8. Combination of resistors in series

Equivalent resistance = R1 + R2 +R3+...

Combination of resistors in parallel

1/Equivalent resistance = 1/R1 + 1/R2 +1/R3+...

Division of current in resistors joined in parallel

i1/i2 = R2/R1

i1 = iR2/(R1 + R2)

9. Batteries connected in series

i = (Є 1 + Є 2)/(R + r0)

Where R = external resistance

r0 = r1 + r2 r1, r2 are internal resistances of two batteries

Batteries connected in parallel

Equivalent emf = Є 0 = [Є 1r2 + Є 2r1]/(r1+r2)

where Є 1, Є 2 are emfs of of batteries , and r1, r2 are internal resistances.

equivalent internal restance = r0 = r1r2/(r1 + r2)

So i in the circuit = ε0/(R + r0)

10. Wheatstone Bridge ; R1 and R2 are two resistances connected in series. R3 and R4 are the other two resistance connected series. If the there is no deflection in the galvanometer

R4 = R3R2/R1

as R1/R2 = R3/R4

11. Charging of the capacitor

q = Є (1 - e^-t/CR)

q is charge on the capacitor, t is time, Є = emf of the battery, C = capacitance, R is resistance of battery and connecting wires,

CR has units of time and is termed time constant. In one time constant τ (=CR) the charge accumulated becomes 0.63 Є C.

12. Discharging of the capacitor

q = Qe^(-t/CR)

where q is charge remaining on the capacitor

Q is the initial charge

In one time constant 0.63% is discharged.

The current at time t = I = lim ∆t→0 ( ∆Q/∆t) = dQ/dt

Q is charge, t is time

**2. Current density**Average current density j (bar) = Δi/ΔA

i is current and A is area of the conductor

The current density at a point P is

j = lim ∆t→0 (Δi/ΔA) = di/dS

If current i is uniformly distributed over an area S and is perpendicular to it

j = i/S

For a finite area i = ∫j.dS

Where

j = density of current (vector)

dS = area (vector)

3. If τ be the average time between successive collisions, the distance drifted during this period is

l = ½ a(τ) = ½ (eE/m)( τ) ²

The drift speed is

v

_{d}= l/ τ = ½ (eE/m)τ = kEτ the average time between successive collisions, is constant for a given material at a given temperature.

Relation between current density and drift speed

j = i/A = nev

_{d}**4. Ohm's law**j = σE

E is field and σ is electrical conductivity of the material.

Resistivity of a material ρ = 1/σ

**Another form of Ohm's law**V = voltage difference between the ends of a conductor = El (l = length of the conductor)

V = Ri

R = resistance of the conductor = ρ*l/A

1/R is called conductance

5. As temperature of a resistor increases its resistance increases. The relation can be expressed as

R(T) = R(T

_{0})[1 + α(T - T_{0})]α is called temperature coefficient of resistivity.

6. The work done by the battery force per unit charge is

Є = W/q = F

_{b}*d/qThis Є is called the emf of the battery. Please note that emf is work done/charge.

If nothing is externally connected

F

_{b}= qE orF

_{b}*d = qEd = qV (because V =Ed)V = potential difference between the terminals

As Є = W/q = F

_{b}*d/q = qV/q = VTherefore Є = V

7. Thermal energy produced in a register

U = i²Rt

Power developed = P = U/t = i²R = Vi

8. Combination of resistors in series

Equivalent resistance = R1 + R2 +R3+...

Combination of resistors in parallel

1/Equivalent resistance = 1/R1 + 1/R2 +1/R3+...

Division of current in resistors joined in parallel

i1/i2 = R2/R1

i1 = iR2/(R1 + R2)

9. Batteries connected in series

i = (Є 1 + Є 2)/(R + r0)

Where R = external resistance

r0 = r1 + r2 r1, r2 are internal resistances of two batteries

Batteries connected in parallel

Equivalent emf = Є 0 = [Є 1r2 + Є 2r1]/(r1+r2)

where Є 1, Є 2 are emfs of of batteries , and r1, r2 are internal resistances.

equivalent internal restance = r0 = r1r2/(r1 + r2)

So i in the circuit = ε0/(R + r0)

10. Wheatstone Bridge ; R1 and R2 are two resistances connected in series. R3 and R4 are the other two resistance connected series. If the there is no deflection in the galvanometer

R4 = R3R2/R1

as R1/R2 = R3/R4

11. Charging of the capacitor

q = Є (1 - e^-t/CR)

q is charge on the capacitor, t is time, Є = emf of the battery, C = capacitance, R is resistance of battery and connecting wires,

CR has units of time and is termed time constant. In one time constant τ (=CR) the charge accumulated becomes 0.63 Є C.

12. Discharging of the capacitor

q = Qe^(-t/CR)

where q is charge remaining on the capacitor

Q is the initial charge

In one time constant 0.63% is discharged.

### IIT JEE Physics Formula Revision 33. Thermal and Chemical Effects of Electric Current

1. Joule's laws of heating

Work done by the electric field on the free electrons in time t is

W = potential difference * charge

= V(it)

= (iR)it = i²Rt

2. If θ

θ

3. Thermo emf depends on temperature with the relation

Є

Where a

This gives dЄ

The quantity dЄ

5. Law of intermediate Metal

Suppose Є

If hot junctions and cold junctions of the three thermocouples are at the same temperature,

Then,

Є

6. Law of intermediate temperature

Let Є

Є

7. The actual emf produced in a thermocouple loop is the algebraic sum of the net Peltier effect and the net Thomspon effect.

So emf in a thermocouple loop = Є

(Π

(More explanation is needed for the formula.)

Work done by the electric field on the free electrons in time t is

W = potential difference * charge

= V(it)

= (iR)it = i²Rt

2. If θ

_{c}, θ_{n}, θ_{i}are temperature of the cold junction, neutral temperature and inversion temperature respectively_{c}θ

_{n}– θ_{c}= θ_{i}- θ_{n}3. Thermo emf depends on temperature with the relation

Є

_{AB}= a_{AB}θ + ½ bsub>ABθ²Where a

_{AB}and b_{AB}are constants for a pair of metals A and B.This gives dЄ

_{AB}/dθ = a_{AB}+ bsub>ABθThe quantity dЄ

_{AB}/dθ is called thermoelectric power at temperature θ.5. Law of intermediate Metal

Suppose Є

_{AB}, Є_{AC}, Є_{BC}are emfs from thermocoupes AB,AC, and BC.If hot junctions and cold junctions of the three thermocouples are at the same temperature,

Then,

Є

_{AB}= Є_{AC}- Є_{BC}6. Law of intermediate temperature

Let Є

_{ θ1, θ2}, represent the thermo-emf of a given thermocouple when the temperatures of junctions are maintained at θ1and θ2. ThenЄ

_{ θ1, θ2}= Є_{ θ1, θ3}+ Є_{ θ3, θ2}7. The actual emf produced in a thermocouple loop is the algebraic sum of the net Peltier effect and the net Thomspon effect.

So emf in a thermocouple loop = Є

_{AB}=(Π

_{AB})_{T}- (Π_{AB})_{T0}+(T – T0) (σ_{A}- σ_{A})(More explanation is needed for the formula.)

### IIT JEE Physics Formula Revision 34. Magnetic Field

1. Magnetic field can be defined mathematically as

Equation uniquely determines the direction of magnetic field B from the rules of the vector product.

q = charge

2. Motion of a Charged particle in a uniform magnetic field

qvB = mv²/r

r = mv/qB

The time taken to complete the circle is

T = 2πr/v = 2πm/qB

Frequency of revolutions is

ν = 1/T = qB/2πm

This frequency is called cyclotron frequency.

6. If a straight wire of length l carry8ng a current i is placed in a uniform magnetic field B, the force on it is

Where

i = current in the conductor

l = vector of length of the conductor

B = magnetic field

7. Formula for Torque on a current loop

If there is a rectangular loop carrying current i in a uniform magnetic field B

then net torque acting on the loop is

Г = iABsin θ

Where i = current in the loop

A = area (magnitude)

B = magnetic field (magnitude)

θ = the angle of inclination of the loop with the plane perpendicular to the plane of magnetic field.

We can also define in terms of vector

i

If there are n turns in the loop, each turn experiences a torque.

The net torque is

**F**= q**v**×**B**Equation uniquely determines the direction of magnetic field B from the rules of the vector product.

**F**= force (vector)q = charge

**v**= velocity (vector)**B**= magnetic field (vector)2. Motion of a Charged particle in a uniform magnetic field

qvB = mv²/r

r = mv/qB

The time taken to complete the circle is

T = 2πr/v = 2πm/qB

Frequency of revolutions is

ν = 1/T = qB/2πm

This frequency is called cyclotron frequency.

6. If a straight wire of length l carry8ng a current i is placed in a uniform magnetic field B, the force on it is

**F**= i**l**×**B**Where

i = current in the conductor

l = vector of length of the conductor

B = magnetic field

7. Formula for Torque on a current loop

If there is a rectangular loop carrying current i in a uniform magnetic field B

then net torque acting on the loop is

Г = iABsin θ

Where i = current in the loop

A = area (magnitude)

B = magnetic field (magnitude)

θ = the angle of inclination of the loop with the plane perpendicular to the plane of magnetic field.

We can also define in terms of vector

**Г**= i**A**×**B**i

**A**can be termed as**μ**the magnetic dipole moment or simply magnetic moment of the current loop.If there are n turns in the loop, each turn experiences a torque.

The net torque is

**Г**= ni**A**×**B****μ**= ni**A**### IIT JEE Physics Formula Revision 35. Magnetic field due to a Current

Note: vectors are shown in bold letters

1. Biot Savart Law

d

where

dB = magnetic field at point P, due to current element dl

c = speed of light

i = current

r = vector joining the current element to the point P.

[1/ ε

Its value is 4 π*10

In terms of µ

d

The magnitude of the field

dB = (µ

where θ is the angle between dl and r.

the direction of the field is perpendicular to the palne containing the current element and the point P according to the rules of the cross product.

4. Magnetic field due to current in a straight wire

B = (µ

θ1 and θ2 and the value of θ corresponding to the lower end and the upper end respectively.

If the point P is on the perpendicular bisector of the straight wire

B = (µ

a = length of the wire

d = distance between the wire and point P (perpendicular distance)

If the wire is very long such that θ1 = 0 and θ2 = π.

The equation is B = (µ

7. Force between two parallel currents

dF/dl = (µ

dF/dl = force per unit length of the wire W2 due to wire W1

i

d = distance between wires kept parallel to each other.

8. Magnetic field in an axial point

B = (µ

If the point is far away from the centre d is very large compared to a

B = (µ

If the magnetic dipole moment of due to the circular conductor with area πa² and current flowing through it is ‘i’, the µ = i πa² and B is

10. Ampere’s law

The circulation of ∫B.dl of the resultant magnetic field along a closed plane curve is equal to µ

The circulation ∫

11. Magnetic field inside a solenoid

B = µ

n = number of turns per unit length along the length of the solenoid.

i = current passing through the solenoid

12. toroid

B = µ

Where

N = total number of turns

i = current in the toroid

r = distance of the point at which field is being calculated from centre of the toroid.

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1. Biot Savart Law

d

**B**= [1/4 π ε_{0}c²]*(i) (**dl*****r**/r³) ... (1)where

dB = magnetic field at point P, due to current element dl

c = speed of light

i = current

r = vector joining the current element to the point P.

[1/ ε

_{0}c²] is written as µ_{0}and is called the permeability of vacuum.Its value is 4 π*10

^{-7 }T-m/AIn terms of µ

_{0}equation (1) becomesd

**B**= (µ_{0}/4 π)*(i) (**dl*****r**/r³) ... (2)The magnitude of the field

dB = (µ

_{0}/4 π)*(idl sin θ/r²)where θ is the angle between dl and r.

the direction of the field is perpendicular to the palne containing the current element and the point P according to the rules of the cross product.

4. Magnetic field due to current in a straight wire

B = (µ

_{0}i /4 πd) (cos θ1 – cos θ2) … (4)θ1 and θ2 and the value of θ corresponding to the lower end and the upper end respectively.

If the point P is on the perpendicular bisector of the straight wire

B = (µ

_{0}i /4 πd) [2a/√(a² +4d²)] … (5)a = length of the wire

d = distance between the wire and point P (perpendicular distance)

If the wire is very long such that θ1 = 0 and θ2 = π.

The equation is B = (µ

_{0}i /2 πd) … (6)7. Force between two parallel currents

dF/dl = (µ

_{0}i_{1}i_{2})/2 πddF/dl = force per unit length of the wire W2 due to wire W1

i

_{1},i_{2}= current in wire W1 and W2 respectivelyd = distance between wires kept parallel to each other.

8. Magnetic field in an axial point

B = (µ

_{0}ia²)/[2(a²+d²)^{3/2}] .. (8)If the point is far away from the centre d is very large compared to a

B = (µ

_{0}ia²)/2d³If the magnetic dipole moment of due to the circular conductor with area πa² and current flowing through it is ‘i’, the µ = i πa² and B is

**B**= (µ_{0}i /4π)(2**µ**/d³) .. (9)10. Ampere’s law

The circulation of ∫B.dl of the resultant magnetic field along a closed plane curve is equal to µ

_{0}times the total current crossing the area bounded by the closed curve provided the electric field inside the loop remains constant.The circulation ∫

**B**.**dl**over closed curve = µ_{0}*I … (10)11. Magnetic field inside a solenoid

B = µ

_{0}nin = number of turns per unit length along the length of the solenoid.

i = current passing through the solenoid

12. toroid

B = µ

_{0}Ni/2πrWhere

N = total number of turns

i = current in the toroid

r = distance of the point at which field is being calculated from centre of the toroid.

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### IIT JEE Physics Formula Revision 36. Permanent Magnets

1. force

A magnetic charge m placed in a magnetic field B experiences a force.

F = mB .. (1)

2. Magnetic field due to magnetic charge

B = (µ

3. Pole strength due to current

m = IA ...(3)

Where

m = pole strength

I = surface current per unit length of the magnet

A = cross sectional of the magnet

4. Magnetic moment of a bar magnet

M = 2ml … (4)

Where

M = Magnetic moment of a bar magnet

M = pole strength

2l = magnetic length of the bar magnet

5. Potential energy at an angle θ

U(θ) = -MB cos θ = -M.B …(5)

6. Magnetic field due to a bar magnet

End on position. A position on the magnetic axis of a bar magnet is called an end on position

B = (µ

If d is very large compared to l, then

B = (µ

8. Broad-on Position

B = (µ

= (µ

If d is very large compared to l,

B = (µ

10. Magnetic scalar potential

V(r2) – V(r1) = -

11. The component of the magnetic field in any direction is given by

B

12.For a pole of pole strength m, the field at a distance r is

B = (µ

So the potential at a distance r is

V( r )= -

= (µ

13. Magnetic scalar potential due to a magnetic dipole

Magnetic scalar potential at a point P which is at a distance r from the mid point of the magnetic dipole, and the angle between the dipole axis and the line joining the mid point of the dipole to the point P is θ

V = (µ

Where

M = 2ml =magnetic moment of the dipole

14. Magnetic field due to dipole

Magnetic field at P =

(µ

17. current in galvanometer

i = K tan θ .. (17)

where K = 2rB

18. Current in moving coil galvanometer

i = (k/nAB) θ … (18)

the constant (k/nAB) is called the galvanometer constant and may be found by passing a known current, measuring the deflection θ and putting these values in equation (18).

19 Shunt

Shunt is a small resistance.

Current in galvanometer

i

i = main current

i

20. tangent law of perpendicular fields

B = B

21. Deflection magnetometer

Tan A position

M/ B

22. Deflection magnetometer Tan B position

M/ B

23. Oscillation Magnetometer

Time period T = 2 π/ ω = 2 π√ (I/M B

From equation (23)

M B

25. Gauss’s law for magnetism: the flux of the magnetic field through any closed surface is zero

integral over the closed surface ∫B.ds = 0 .. (25)

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A magnetic charge m placed in a magnetic field B experiences a force.

F = mB .. (1)

2. Magnetic field due to magnetic charge

B = (µ

_{0}/4 π)(m/r²)3. Pole strength due to current

m = IA ...(3)

Where

m = pole strength

I = surface current per unit length of the magnet

A = cross sectional of the magnet

4. Magnetic moment of a bar magnet

M = 2ml … (4)

Where

M = Magnetic moment of a bar magnet

M = pole strength

2l = magnetic length of the bar magnet

5. Potential energy at an angle θ

U(θ) = -MB cos θ = -M.B …(5)

6. Magnetic field due to a bar magnet

End on position. A position on the magnetic axis of a bar magnet is called an end on position

B = (µ

_{0}/4 π)[(2Md)/(d² – l²)²] …(6)If d is very large compared to l, then

B = (µ

_{0}/4 π)(2M/d³) … (7)8. Broad-on Position

B = (µ

_{0}/4 π)[m 2l/(d² + l²)^{3/2 }] …(8)= (µ

_{0}/4 π)[M/(d² + l²)^{3/2 }]If d is very large compared to l,

B = (µ

_{0}/4 π)[M/(d³)10. Magnetic scalar potential

V(r2) – V(r1) = -

_{r1}∫^{r2 }B.dr … (10)11. The component of the magnetic field in any direction is given by

B

_{l}= -dV/dl ... (11)12.For a pole of pole strength m, the field at a distance r is

B = (µ

_{0}/4 π)[m/r²)So the potential at a distance r is

V( r )= -

_{∞}∫^{r }(µ_{0}/4 π)[m/r²)dr= (µ

_{0}/4 π)[m/r) ... (12)13. Magnetic scalar potential due to a magnetic dipole

Magnetic scalar potential at a point P which is at a distance r from the mid point of the magnetic dipole, and the angle between the dipole axis and the line joining the mid point of the dipole to the point P is θ

V = (µ

_{0}/4 π)[Mcos θ /r²)Where

M = 2ml =magnetic moment of the dipole

14. Magnetic field due to dipole

Magnetic field at P =

(µ

_{0}/4 π)[M /r²)√(1 +3 cos² θ)] .. (14)17. current in galvanometer

i = K tan θ .. (17)

where K = 2rB

_{H}/µ_{0}for the given galvanometer at a given place.18. Current in moving coil galvanometer

i = (k/nAB) θ … (18)

the constant (k/nAB) is called the galvanometer constant and may be found by passing a known current, measuring the deflection θ and putting these values in equation (18).

19 Shunt

Shunt is a small resistance.

Current in galvanometer

i

_{g}= [R_{s}/(R_{s}+_{g})]*ii = main current

i

_{g}= current that goes through galvanometer20. tangent law of perpendicular fields

B = B

_{H}tan θ .. (20)21. Deflection magnetometer

Tan A position

M/ B

_{H}= (4 π/µ_{0})[(d² – l²)²/2d] tan θ22. Deflection magnetometer Tan B position

M/ B

_{H}= (4 π/µ_{0})[(d² – l²)^{3/2}] tan θ.. (22)23. Oscillation Magnetometer

Time period T = 2 π/ ω = 2 π√ (I/M B

_{H}) .. (23)From equation (23)

M B

_{H}) = 4 π²I/T² … (24)25. Gauss’s law for magnetism: the flux of the magnetic field through any closed surface is zero

integral over the closed surface ∫B.ds = 0 .. (25)

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### IIT JEE Physics Formula Revision 37. Magnetic Properties of Matter

Ch. 37 Magnetic Properties of Matter – Formulae

1. Magnetization vector = Magnetic moment per unit volume

2. Magnetic intensity

Where

Magnetic intensity due to a magnetic pole of pole strength m at a distance r from it is

H = m/(4 πr²) …(5)

6. Magnetic susceptibility

Χ is called the susceptibility of the material.

7. Permeability

µ = µ

µ

µ

9. Curie’s law

χ = c/T … (9)

where c = Curie’s constant

For ferromagnetic materials

Χ = c’/(T - T

Where

T

c’ = constant

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1. Magnetization vector = Magnetic moment per unit volume

**I**=**M**/V2. Magnetic intensity

**H**=**B**/µ_{0}-**I**.. (2)Where

**H**= magnetic intensity**B**- resultant magnetic field**I**= intensity of magnetizationMagnetic intensity due to a magnetic pole of pole strength m at a distance r from it is

H = m/(4 πr²) …(5)

6. Magnetic susceptibility

**I**= χ**H**… (6)Χ is called the susceptibility of the material.

7. Permeability

**B**= µ**H**… (7)µ = µ

_{0}(1+χ) is a constant and is called the permeability of the material.µ

_{0}is the permeability of vacuum.µ

_{r}= µ/µ_{0}= 1+ χ is called the relative permeability of the material.9. Curie’s law

χ = c/T … (9)

where c = Curie’s constant

For ferromagnetic materials

Χ = c’/(T - T

_{c})Where

T

_{c}is the Curie point andc’ = constant

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### IIT JEE Physics Formula Revision 38. Electro Magentic Induction

1. Faraday’s law of electromagnetic induction

Є = -dФ/dt … (1)

Where

Є = emf produced

Ф = ∫

2. i = Є/R = -(1/R) dФ/dt …(2)

where i = current in the circuit

R = resistance of the circuit

3. Є = vBl

Where

Є = emf produced

v = velocity of the conductor

B = magnetic field in which the conductor is moving

l = length of the conductor

4. Induced electric field

∫

where

E = induced electric filed due to magnetic field B

5. Self induction

Magnetic field through the area bounded by a current-carrying loop is proportional to the current flowing through it.

Ф = Li … (5)

Where

Ф = ∫

L = is a constant called the self-inductance of the loop.

i = current through the loop.

6. Self induced EMF

Є = -dФ/dt = -Ldi/dt ….(6)

7. Self inductance of a long solenoid

L = µ

8. Growth of current through an LR circuit

i = i

= i

where

i = current in the circuit at time t

i

Є = applied emf

R = resistance of the circuit

L = inductance of the circuit

τ = L/R = time constant of the LR circuit

10. Decay of current in a LR circuit

i = i

= i

i = current in the circuit at time t

i

R = resistance of the circuit

L = inductance of the circuit

τ = L/R = time constant of the LR circuit

12. Energy stored in an inductor

U = ½ Li² … (12)

13. Energy density

u = U/V = B²/2µ

14. Mutual induction

Ф = Mi … (14)

Where

M = constant called mutual inductance of the given pair of circuits

Є = -Mdi/dt …. (15)

Є = -dФ/dt … (1)

Where

Є = emf produced

Ф = ∫

**B**.**dS**= the flux of the magnetic field through the area.2. i = Є/R = -(1/R) dФ/dt …(2)

where i = current in the circuit

R = resistance of the circuit

3. Є = vBl

Where

Є = emf produced

v = velocity of the conductor

B = magnetic field in which the conductor is moving

l = length of the conductor

4. Induced electric field

∫

**E**.**dl**= -dФ/dt .. (4)where

E = induced electric filed due to magnetic field B

5. Self induction

Magnetic field through the area bounded by a current-carrying loop is proportional to the current flowing through it.

Ф = Li … (5)

Where

Ф = ∫

**B**.**dS**= the flux of the magnetic field through the area.L = is a constant called the self-inductance of the loop.

i = current through the loop.

6. Self induced EMF

Є = -dФ/dt = -Ldi/dt ….(6)

7. Self inductance of a long solenoid

L = µ

_{0}n² πr² l … (7)8. Growth of current through an LR circuit

i = i

_{0}(1 - e^{-tR/L}) … (8)= i

_{0}(1 - e^{-t/ τ }) … (9)where

i = current in the circuit at time t

i

_{0}= Є/RЄ = applied emf

R = resistance of the circuit

L = inductance of the circuit

τ = L/R = time constant of the LR circuit

10. Decay of current in a LR circuit

i = i

_{0}(1 - e^{-tR/L}) … (10)= i

_{0}(1 - e^{-t/ τ }) … (11)i = current in the circuit at time t

i

_{0}= current in the circuit at time t = 0R = resistance of the circuit

L = inductance of the circuit

τ = L/R = time constant of the LR circuit

12. Energy stored in an inductor

U = ½ Li² … (12)

13. Energy density

u = U/V = B²/2µ

_{0}14. Mutual induction

Ф = Mi … (14)

Where

M = constant called mutual inductance of the given pair of circuits

Є = -Mdi/dt …. (15)

### IIT JEE Physics Formula Revision 39. Alternating current

Ch. 39 Alternating Current

1. Alternating current

i = i

where

i = current at time t

i

current repeats after each time interval T = 2 π/ω

2. Emf produced in a AC generator

Є = NBA ω sin ωt = Є

Where

N = number of turns on the armature

B = Magnetic field

A = area of the coil on the armature

ω = angular velocity of the armature

3. RMS (rms) current in a AC circuit

i

The alternating current and voltage are generally measured and expressed in termsof their rms values. When the household current is expressed as 220 V AC, it means that the rms value of voltage is 220 V. The peak voltage value is (220 V) √2 = 311 V.

4. current in an AC circuit containing only resistor

i = i

where

i

6. Ac circuit containing only a capacitor

i = i

where

i

= Є

8. AC circuit containing only an inductor

i = (Є

or i = i

where i

10. Impedance

The peak current and the peak emf in Ac circuits may be written as

i

where Z = R for a purely resistive circuit

Z = 1/ ωC for a purely capacitive circuit

Z = ωL for a purely inductive circuit

11. Vector method to find the current in an AC circuit

i = (Є

where

Z = √(R² + X²)

R = resistance in the circuit

X = reactance in the circuit (X due to capacitance is 1/ ωC and X due to inductance is ωL)

φ is to be determined from tan φ = X/R

15. Power in Ac Circuits

P = Є

16. Voltages in primary and secondary of transformer

Є

Power transfer

Є

i

‘-‘ sign shows that i

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1. Alternating current

i = i

_{0}sin (ωt + φ) .. (1)where

i = current at time t

i

_{0}= peak currentcurrent repeats after each time interval T = 2 π/ω

2. Emf produced in a AC generator

Є = NBA ω sin ωt = Є

_{0}sin ωt … (2)Where

N = number of turns on the armature

B = Magnetic field

A = area of the coil on the armature

ω = angular velocity of the armature

3. RMS (rms) current in a AC circuit

i

_{rms}= i_{0}/√2 … (3)The alternating current and voltage are generally measured and expressed in termsof their rms values. When the household current is expressed as 220 V AC, it means that the rms value of voltage is 220 V. The peak voltage value is (220 V) √2 = 311 V.

4. current in an AC circuit containing only resistor

i = i

_{0}sin ωt …… (4)where

i

_{0}= Є_{0}/R ... (5)6. Ac circuit containing only a capacitor

i = i

_{0}cos ωt … (6)where

i

_{0}=C Є_{0}ω= Є

_{0}/(1/ωC) … (7)8. AC circuit containing only an inductor

i = (Є

_{0}/ωL) sin (ωt – π/2) …(8)or i = i

_{0}sin (ωt – π/2)where i

_{0}= Є_{0}/ωL10. Impedance

The peak current and the peak emf in Ac circuits may be written as

i

_{0}= Є_{0}/Z … (10)where Z = R for a purely resistive circuit

Z = 1/ ωC for a purely capacitive circuit

Z = ωL for a purely inductive circuit

11. Vector method to find the current in an AC circuit

i = (Є

_{0}/Z) sin (ωt + φ)where

Z = √(R² + X²)

R = resistance in the circuit

X = reactance in the circuit (X due to capacitance is 1/ ωC and X due to inductance is ωL)

φ is to be determined from tan φ = X/R

15. Power in Ac Circuits

P = Є

_{rms}i_{rms}cos φ16. Voltages in primary and secondary of transformer

Є

_{2}= - (N_{2}/ N_{1}) Є_{1}.... (16)Power transfer

Є

_{1}i_{1}= Є_{2}i_{2}…(17)i

_{2}= - (N_{1}/ N_{2}) i_{1}‘-‘ sign shows that i

_{2}is 180° out of phase with i_{1}.Join

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### IIT JEE Physics Formula Revision 40. Electromagentic Waves

Electromagnetic waves formula revision

1. The wave equation for light propagating in x-direction in vacuum may be written as

E = E

Where E is the sinusoidally varying electric field at the position x at time t.

c is the speed of light in vacuum.

The electric field is in the Y-Z plane. It is perpendicular to the direction of propagation of the wave.

There is a sinusoidally varying magnetic field associated with the electric field when light is propagating. This magnetic field is perpendicular to the direction of wave propagation and the electric field E.

B = B

Such a combination of mutually perpendicular electric and magnetic fields constitute an electromagnetic wave in vacuum.

2. Maxwell generalised Ampere’s law to

∫B.dl = µ

i

Where

Φ

3. Maxwell’s Equations

Gauss’s laws for electricity and magnetism, Faraday’s law and Ampere’s are collectively known as Maxwell’s equations

Gauss’s law of electricity

∮E.ds = q/ ε

Gauss’s law for magnetism

∮B.ds = 0

Faraday’s law

∮E.dl = -dΦ

Ampere’s law

∮B.dl = µ

i

These equations are satisfied by a plane electromagnetic wave given by

E

B

4. c = i/√(µ

the value calculated from this expression comes out to be 2.99793*10^8 m/s which was same as the experimentally measured value of speed of light in vacuum. This also provides a confirmatory proof that light is an electromagnetic wave.

5. Total energy of the electromagnetic wave is

U = ½ ε

When we substitute the values of E and B in the above equation and take an average over a longer period of time

u

6. intensity of electromagnetic wave is per unit area per unit time I = U/AΔt = u

In terms of maximum electric field (substituting the value of u

I = ½ ε

7. The electromagnetic wave also carries linear momentum with it. The linear momentum carried by the portion of wave having energy U is given by

p = U/c

where U = energy contained in the portion of the wave.

1. The wave equation for light propagating in x-direction in vacuum may be written as

E = E

_{0}sin ω(t-x/c)Where E is the sinusoidally varying electric field at the position x at time t.

c is the speed of light in vacuum.

The electric field is in the Y-Z plane. It is perpendicular to the direction of propagation of the wave.

There is a sinusoidally varying magnetic field associated with the electric field when light is propagating. This magnetic field is perpendicular to the direction of wave propagation and the electric field E.

B = B

_{0}sin ω(t-x/c)Such a combination of mutually perpendicular electric and magnetic fields constitute an electromagnetic wave in vacuum.

2. Maxwell generalised Ampere’s law to

∫B.dl = µ

_{0}(i + i_{d})i

_{d}= ε_{0}*(d Φ_{E}/dt)Where

Φ

_{E}/ = the flux of the electric field through the area bounded by the closed curve along which the circulation of B is calculated.3. Maxwell’s Equations

Gauss’s laws for electricity and magnetism, Faraday’s law and Ampere’s are collectively known as Maxwell’s equations

Gauss’s law of electricity

∮E.ds = q/ ε

_{0}Gauss’s law for magnetism

∮B.ds = 0

Faraday’s law

∮E.dl = -dΦ

_{B}/dtAmpere’s law

∮B.dl = µ

_{0}(i + i_{d})i

_{d}= ε_{0}*(d Φ_{E}/dt)These equations are satisfied by a plane electromagnetic wave given by

E

_{y}= E = E_{0}sin ω(t-x/c)B

_{z}= B = B_{0}sin ω(t-x/c)4. c = i/√(µ

_{0}ε_{0})the value calculated from this expression comes out to be 2.99793*10^8 m/s which was same as the experimentally measured value of speed of light in vacuum. This also provides a confirmatory proof that light is an electromagnetic wave.

5. Total energy of the electromagnetic wave is

U = ½ ε

_{0}E²dV + B²dV/2µ_{0}When we substitute the values of E and B in the above equation and take an average over a longer period of time

u

_{av}= ½ ε_{0}E_{0}² = B_{0}²/2µ_{0}6. intensity of electromagnetic wave is per unit area per unit time I = U/AΔt = u

_{av}c.In terms of maximum electric field (substituting the value of u

_{av}c)I = ½ ε

_{0}E_{0}²c7. The electromagnetic wave also carries linear momentum with it. The linear momentum carried by the portion of wave having energy U is given by

p = U/c

where U = energy contained in the portion of the wave.

## Thursday, February 7, 2008

### IIT JEE Physics Formula Revision 41. Electric Current through Gases

1.

Sparking potential of a gas in a discharge tube is a function of the product of the pressure of the gas and the separation between the electrodes.

V = f(pd)

2. Experiment to Determine e/m by Thomson

Cathode rays are subjected to electric and magnetic fields.

If both the electric field and magnetic field are switched on and the values are so chosen that

v = E/B

The magnetic field evB will exactly cancel the electric force eE and the beam will pass undeflected. If the potential difference between the anode and the cathode is V, the speed of the electrons coming out of A is given by

½ mv² = eV

as v = E/B

½ m (e/B) ² = eV

therefore e/m = E²/2B²V

3. If n thermions are ejected per unit time by a metal surface, the thermionic current i = ne. This current is given by the Richardson-Dushman equation.

i = ne = AST²e

where

S = surface area

T = the absolute temperature of the surface.

φ = work function of the metal

K = Boltzmann constant

A = constant which depends only the nature of the metal

**Paschen's law**Sparking potential of a gas in a discharge tube is a function of the product of the pressure of the gas and the separation between the electrodes.

V = f(pd)

2. Experiment to Determine e/m by Thomson

Cathode rays are subjected to electric and magnetic fields.

If both the electric field and magnetic field are switched on and the values are so chosen that

v = E/B

The magnetic field evB will exactly cancel the electric force eE and the beam will pass undeflected. If the potential difference between the anode and the cathode is V, the speed of the electrons coming out of A is given by

½ mv² = eV

as v = E/B

½ m (e/B) ² = eV

therefore e/m = E²/2B²V

3. If n thermions are ejected per unit time by a metal surface, the thermionic current i = ne. This current is given by the Richardson-Dushman equation.

i = ne = AST²e

^{- φ/kT}where

S = surface area

T = the absolute temperature of the surface.

φ = work function of the metal

K = Boltzmann constant

A = constant which depends only the nature of the metal

### IIT JEE Physics Formula Revision 42. Photoelectric Effect and Waveparticle Duality

1. Relation between properties of photon and properties of light waves.

E and p are energy and linear momentum of a photon of light.

ν and λ are the frequency and wavelength of the same light when it is considered (behaves) as a wave.

Then E = hν = hc/λ

p = h/λ = E/c ... (42.1)

wherein h is a universal constant known as the Planck constant and has a value 6.626*10^-34 J-s and is also equal to 4.136*10^-15 eV-s.

C = velocity of light vacuum = 299,792,458 m/s ≈ 3.0*10^8 m/s

2. The maximum kinetic energy of the electron that comes out due to energy E supplied is:

Kmax = E – φ

Some energy from the E – φ is dissipated as the electron may have some collision before escaping from the material.

If monochromatic light of wave length is incident on the metal surface, photons of energy hc/ λ fall on the surface. The maximum kinetic energy of an electrons that comes out due to these photons is:

Kmax = hc/λ - φ = h υ - φ

The above equation is called Einstein's photoelectric equation.

3. Writing work function φ as hυ

υ λ = c

υ

λ

K

4. Relation between Maximum kinetic energy of photoelectrons and stopping potential:

As a photoelectron travels from the cathode to the anode, the potential energy increases by eV

The maximum kinetic energy a photoelectron will have is hc/λ - φ

Hence eV

V

5. A relation for wavelength of electron was proposed by Louis Victor de Broglie.

The proposed expression for wavelength is

λ = h/p

Where p is the momentum of the electron and

h is the Planck constant.

E and p are energy and linear momentum of a photon of light.

ν and λ are the frequency and wavelength of the same light when it is considered (behaves) as a wave.

Then E = hν = hc/λ

p = h/λ = E/c ... (42.1)

wherein h is a universal constant known as the Planck constant and has a value 6.626*10^-34 J-s and is also equal to 4.136*10^-15 eV-s.

C = velocity of light vacuum = 299,792,458 m/s ≈ 3.0*10^8 m/s

2. The maximum kinetic energy of the electron that comes out due to energy E supplied is:

Kmax = E – φ

Some energy from the E – φ is dissipated as the electron may have some collision before escaping from the material.

If monochromatic light of wave length is incident on the metal surface, photons of energy hc/ λ fall on the surface. The maximum kinetic energy of an electrons that comes out due to these photons is:

Kmax = hc/λ - φ = h υ - φ

The above equation is called Einstein's photoelectric equation.

3. Writing work function φ as hυ

_{0}(h multiplied by frequency)υ λ = c

υ

_{0}= c/ λ_{0}λ

_{0}= threshold wavelengthK

_{max}= h(υ - υ_{0})4. Relation between Maximum kinetic energy of photoelectrons and stopping potential:

As a photoelectron travels from the cathode to the anode, the potential energy increases by eV

_{0}. This is equal to the decrease in the kinetic energy of the photoelectron.The maximum kinetic energy a photoelectron will have is hc/λ - φ

Hence eV

_{0}. = hc/λ – φV

_{0}= hc/e(1/ λ) – φ/e5. A relation for wavelength of electron was proposed by Louis Victor de Broglie.

The proposed expression for wavelength is

λ = h/p

Where p is the momentum of the electron and

h is the Planck constant.

### IIT JEE Physics Formula Revision 43. Bohr's Model and Physics of Atom

1. Equation for wavelengths of radiation emitted by the hydrogen atom.

1/λ = R [1/n² - 1/m²]

where R = 1.09737*10^7 m

n and m are integers with m>n.

2. Velocity of an electron when it is in a stationary orbit represented by n which is an integer

v = Ze²/2 ε

3. Radius of a stationary orbit based on n which is an integer

r = ε

4. Kinetic energy when electron is in nth orbit is

K = ½ mv² = mZ²e

5. The potential energy of the atom is

V = - Ze²/4π ε

The expression for potential energy is obtained by assuming the potential energy to be zero when the nucleus and the electron are widely separated.

6. The total energy of the atom is

E = K+V = - mZ²e

7. For a hydrogen like ion with Z protons in the nucleus,

r

8. Energy of hydrogen atom when the single electron is in the nth orbit.

En = E1/n² = -13.6/n² eV

-13.6 eV is the energy when the electron is in the n = 1 orbit.

Note that the energy is expressed in negative units, so that larger magnitude means lower energy.

9. If an electron jumps from mth orbit to nth orbit (m>n) of a hydrogen like ion, the energy of the atom gets reduced from Em to En. The wavelength of the emitted radiation will be

1/ λ = (Em – En)/hc = RZ²{1/n² - 1/m²]

10. The wave function of the electron ψ(r,t) is obtained from the Schrodinger’s equation

-(h²/8π²m) [∂²ψ /∂x² + ∂²ψ /∂y² + ∂²ψ/∂z²] - Ze²ψ/4πε

where

(x.y,z ) refers to a point with the nucleus as the origin and r is the distance of this point from the nucleus.

E refers to the energy.

Z is the number of protons.

11. The energy of the wave function of characterized by n,l, and m

En = - mZ²e

12. When n = 1, the wave function of the hydrogen atom is

ψ(r,) = ψ

where

ψ

a

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1/λ = R [1/n² - 1/m²]

where R = 1.09737*10^7 m

^{-1}.n and m are integers with m>n.

2. Velocity of an electron when it is in a stationary orbit represented by n which is an integer

v = Ze²/2 ε

_{0}hn3. Radius of a stationary orbit based on n which is an integer

r = ε

_{0}h²n²/πmZe²4. Kinetic energy when electron is in nth orbit is

K = ½ mv² = mZ²e

^{4}/8 ε_{0}²h²n²5. The potential energy of the atom is

V = - Ze²/4π ε

_{0}r = -mZ²e^{4}/4ε_{0}²h²n²The expression for potential energy is obtained by assuming the potential energy to be zero when the nucleus and the electron are widely separated.

6. The total energy of the atom is

E = K+V = - mZ²e

^{4}/8 ε_{0}²h²n²7. For a hydrogen like ion with Z protons in the nucleus,

r

_{n}= radius of ‘n’ th orbit = n²a_{0}/Z8. Energy of hydrogen atom when the single electron is in the nth orbit.

En = E1/n² = -13.6/n² eV

-13.6 eV is the energy when the electron is in the n = 1 orbit.

Note that the energy is expressed in negative units, so that larger magnitude means lower energy.

9. If an electron jumps from mth orbit to nth orbit (m>n) of a hydrogen like ion, the energy of the atom gets reduced from Em to En. The wavelength of the emitted radiation will be

1/ λ = (Em – En)/hc = RZ²{1/n² - 1/m²]

10. The wave function of the electron ψ(r,t) is obtained from the Schrodinger’s equation

-(h²/8π²m) [∂²ψ /∂x² + ∂²ψ /∂y² + ∂²ψ/∂z²] - Ze²ψ/4πε

_{0}r = E ψwhere

(x.y,z ) refers to a point with the nucleus as the origin and r is the distance of this point from the nucleus.

E refers to the energy.

Z is the number of protons.

11. The energy of the wave function of characterized by n,l, and m

_{l}depends only on n and may be written asEn = - mZ²e

^{4}/8 ε_{0}²h²n²12. When n = 1, the wave function of the hydrogen atom is

ψ(r,) = ψ

_{100}= √(Z³/ π a_{0}²) *(e^{-r/ a0})where

ψ

_{100}denotes that n =1, l = 0 and m_{l}= 0a

_{0}= Bohr radiusJoin Orkut community

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### IIT JEE Physics Formula Revision 44. X-Rays

1. K = eV

where K = Kinetic energy of an electron when it hits the target

V = potential difference applied between the target and the filament

2. λ = hc/E

Where

λ = wave length of the X-ray

E = kinteic energy of the electron due to which the X-ray is emitted

3. λ

λ

V = potential difference applied between the target and the filament

4. Moseley's law

Square root of frequency of X rays = a(Z-b)

√(ν) = a(Z-b)

where

ν = frequency of X-rays

Z = position number of element.

a and b are constants

5. Bragg’s law

2d sin θ = n λ

d = interplanar spacing of the crystal on which X-rays are incident

θ = is the incident angle at which X-rays are strongly reflected.

n = 1,2,3 …

λ = wave length of X-rays

where K = Kinetic energy of an electron when it hits the target

V = potential difference applied between the target and the filament

2. λ = hc/E

Where

λ = wave length of the X-ray

E = kinteic energy of the electron due to which the X-ray is emitted

3. λ

_{min}= hc/eVλ

_{min}= cutoff wavelength below which no X-rays are emittedV = potential difference applied between the target and the filament

4. Moseley's law

Square root of frequency of X rays = a(Z-b)

√(ν) = a(Z-b)

where

ν = frequency of X-rays

Z = position number of element.

a and b are constants

5. Bragg’s law

2d sin θ = n λ

d = interplanar spacing of the crystal on which X-rays are incident

θ = is the incident angle at which X-rays are strongly reflected.

n = 1,2,3 …

λ = wave length of X-rays

### IIT JEE Physics Formula Revision 46. Nucleus

1. R = R

where R

2. B = [Zm{

– m{

Where

B = binding energy of the nucleus

m{

m{

3. Binding energy per nucleon = B/A = a

4. Mass excess =

(mass of atom – A’)c²

5. Mass excess = 931(m-A)MeV

6. Alpha decay process is represented by

7. q value of the process

Q = [m{

8. Beta decay process

N --> p + e + antineutrino

9. Beta decay proces

e is also shown as beta minus.

10. kinetic energy Q available ot the product particles is

Q = [m(

11. Proton conversion process – Beta plus decay

P --> n + e

12.

13.Q value of the decay

Q = {m(

_{0}A^(1/3) .. (46.1)where R

_{0}= 1.1*10^-15 m ≈ 1.1 fm and A is the mass number2. B = [Zm{

_{1}^{1}H} +Nm_{n}– m{

_{Z}^{Z+n}}]c²Where

B = binding energy of the nucleus

m{

_{1}^{1}H} is the mass of a hydrogen atomm{

_{Z}^{Z+n}} is the mass of an atom with Z protons and N neutrons3. Binding energy per nucleon = B/A = a

_{1}- a_{2}/(A^{1/3}) - a_{3}Z(Z-1)/ (A^{4/3})4. Mass excess =

(mass of atom – A’)c²

5. Mass excess = 931(m-A)MeV

6. Alpha decay process is represented by

_{Z}^{A}X -->_{Z-2}^{A-4}Y +_{2}^{4}He7. q value of the process

Q = [m{

_{Z}^{A}X} – m{_{Z-2}^{A-4}Y} – m{_{2}^{4}He}]c²8. Beta decay process

N --> p + e + antineutrino

9. Beta decay proces

_{Z}^{A}X -->_{Z+1}^{A}Y +e + antineutrinoe is also shown as beta minus.

10. kinetic energy Q available ot the product particles is

Q = [m(

_{Z}^{A}X – m{_{Z+1}^{A}Y}]c²11. Proton conversion process – Beta plus decay

P --> n + e

^{+}+ v (neutrino)12.

_{Z}^{A}X -->_{Z-1}^{A}Y + e^{+}+ v (neutrino)13.Q value of the decay

Q = {m(

_{Z}^{A}X) – m(_{Z-1}^{A}Y) – 2m_{e}]c²### IIT JEE Study of Physics

As I have brought my Chemistry study to some stage, I started looking at physics once again. I am trying to write down formulas of each chapter in a notebook first. That way I read the chapter fully. It may take two days for each chapter. I may start the posting the formulas for revision purpose in this blog. Along with it I plan to go through problems/questions of past JEEs and update the study guides with them.

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