1. vav = s/(t2 - t1)
where vav = average speed
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. vav = (r2 - r1)/(t2-t1)
Where
vav = average velocity
r2, r1 = position vectors of a particle
t2, t1 = time instants
5. v = lim Δt→0 Δr/Δt = dr/dt
6. v = |dr|/dt = ds/dt
Where v = magnitude of velocity
s = displacement, for small time intervals magnitude of displacement will be equal to distance.
7. aav = (v2 - v1)/(t2-t1)
8. a = lim Δt→0 Δv/Δt = dv/dt
For Motion in straight line
9. Velocity is v = dx/dt
10. a = dv/dt
11. acceleration is a = dv/dt = d²x/dt²
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. 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)
Showing posts with label Formula revision. Show all posts
Showing posts with label Formula revision. Show all posts
Friday, February 8, 2008
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 = Nf = Mg
Where
M = mass of the object
g = acceleration due to gravity
2. fk = µk Nf
where
fk = magnitude of kinetic friction
µk = coefficient of kinetic friction
3. fmax = µs Nf
where
fmax = maximum static friction
µs = coefficient of static friction
The actual static friction can be less than maximum static friction if the force applied is less than fmax .
4. On an adjustable inclined plane, a block is kept and the angle is gradually increased so that the block begins to move.
Then fmax = mg sin θ
Nf = mg cos θ
There coefficient of static friction µs
µs = fmax / Nf
= 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
µk = coefficient of kinetic friction = tan θ’ = h’/d’
(h is going to decrease and d is going to increase)
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Where
M = mass of the object
g = acceleration due to gravity
2. fk = µk Nf
where
fk = magnitude of kinetic friction
µk = coefficient of kinetic friction
3. fmax = µs Nf
where
fmax = maximum static friction
µs = coefficient of static friction
The actual static friction can be less than maximum static friction if the force applied is less than fmax .
4. On an adjustable inclined plane, a block is kept and the angle is gradually increased so that the block begins to move.
Then fmax = mg sin θ
Nf = mg cos θ
There coefficient of static friction µs
µs = fmax / Nf
= 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
µk = coefficient of kinetic friction = tan θ’ = h’/d’
(h is going to decrease and d is going to increase)
For further interaction join orkut community
IIT-JEE-Academy
http://www.orkut.co.in/Community.aspx?cmm=39291603
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 = ω0t + ½ αt²
ω = ω0 + αt
ω² = ω0² + 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
at = rate of change of speed of the particle in circular motion
at = 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)
er = unit vector along the radius at a point on the circle
er = i cos θ+j sin θ
et = unit vector along the tangent at a point on the circle
et = - 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 er + dv/dt et
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 er
Acceleration of the particle is in the direction of - er , i.e, towards the centre of the circle. The magnitude of the acceleration is
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 er + dv/dt et
the radial component is ar = - ω²r = -v²/r
the tangential component is at = dv/dt
The magnitude of the acceleration
a = √ (ar² + at²)
= √[( 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 = fs/M
Or M v²/r ≤ fs
As fs≤ μsMg
M v²/r ≤ μsMg
That means v²/gr ≤ μs
v≤√( μsgr)
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+ ω4R²sin² 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
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 = ω0t + ½ αt²
ω = ω0 + αt
ω² = ω0² + 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
at = rate of change of speed of the particle in circular motion
at = 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)
er = unit vector along the radius at a point on the circle
er = i cos θ+j sin θ
et = unit vector along the tangent at a point on the circle
et = - 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 er + dv/dt et
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 er
Acceleration of the particle is in the direction of - er , i.e, towards the centre of the circle. The magnitude of the acceleration is
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 er + dv/dt et
the radial component is ar = - ω²r = -v²/r
the tangential component is at = dv/dt
The magnitude of the acceleration
a = √ (ar² + at²)
= √[( 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 = fs/M
Or M v²/r ≤ fs
As fs≤ μsMg
M v²/r ≤ μsMg
That means v²/gr ≤ μs
v≤√( μsgr)
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+ ω4R²sin² 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 vrms
Equation 1 can be written as p = 1/3*ρ*v²rms
So vrms = 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]3/2v²e-(mv²/2kT)dv
14. The speed at which dN/dv is maximum is called the most probable speed.
vp = 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.
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 vrms
Equation 1 can be written as p = 1/3*ρ*v²rms
So vrms = 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]3/2v²e-(mv²/2kT)dv
14. The speed at which dN/dv is maximum is called the most probable speed.
vp = 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. pVγ = constant
γ = Cp/Cv
10. In a reversible adiabatic process
Tγ/pγ-1 = constant
11. In a reversible adiabatic process
TVγ-1 = constant
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. pVγ = constant
γ = Cp/Cv
10. In a reversible adiabatic process
Tγ/pγ-1 = constant
11. In a reversible adiabatic process
TVγ-1 = constant
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
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
λmT = b = constant
6. 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 = σAT4
Where
σ = Stefan Boltzmann constant = 5.67*10-8 W/m²-K4
The energy of thermal radiation emitted by per unit time by a body (not black) of surface area A is given by
U = eσAT4
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 T0
u = eσA(T4 - T04)
8. Newton’s law of cooling
dT/dt = -bA(T-T0)
b = a constant depends on the nature of the surface involved
a = surface area exposed of the body
T- T0) = temperature difference between the body and surrounding
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
λmT = b = constant
6. 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 = σAT4
Where
σ = Stefan Boltzmann constant = 5.67*10-8 W/m²-K4
The energy of thermal radiation emitted by per unit time by a body (not black) of surface area A is given by
U = eσAT4
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 T0
u = eσA(T4 - T04)
8. Newton’s law of cooling
dT/dt = -bA(T-T0)
b = a constant depends on the nature of the surface involved
a = surface area exposed of the body
T- T0) = 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 1/4πε0.
The constant ε0 is called the permittivity of the space and its value is
ε0 = 8.85419*10^-12 C²/N-m²
The intensity of field is defined as Vector E = Vector F/q
(3) E = F/q
Electric field due to a point charge (Q)
(4) E = F/q = Q/4πε0r²
(5) U(r2) - U(r1) = -W = [q1*q2/4πε0]* [(1/r2) - (1/r1)]
Choosing potential energy at infinite separation as zero
(6) U(r) = U(r) - U(∞) = [q1*q2/4πε0]* (1/r)
= q1*q2/4πε0r
(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/4πε0r
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 = -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 = - [1/4πε0][q/(r + (dcos θ)/2)]
Potential at P due to charge q = [1/4πε0][q/(r - (dcos θ)/2)]
Net potential due to q and –q = [1/4πε0](qd cos θ)/r²
17. Electric field due to a dipole
Er = [1/4πε0](2p cos θ)/r³
Eθ = [1/4πε0](p sin θ)/r³
Resultant electric field at P = E= √ (Er²+ Eθ²)
= [1/4πε0](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
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 1/4πε0.
The constant ε0 is called the permittivity of the space and its value is
ε0 = 8.85419*10^-12 C²/N-m²
The intensity of field is defined as Vector E = Vector F/q
(3) E = F/q
Electric field due to a point charge (Q)
(4) E = F/q = Q/4πε0r²
(5) U(r2) - U(r1) = -W = [q1*q2/4πε0]* [(1/r2) - (1/r1)]
Choosing potential energy at infinite separation as zero
(6) U(r) = U(r) - U(∞) = [q1*q2/4πε0]* (1/r)
= q1*q2/4πε0r
(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/4πε0r
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 = -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 = - [1/4πε0][q/(r + (dcos θ)/2)]
Potential at P due to charge q = [1/4πε0][q/(r - (dcos θ)/2)]
Net potential due to q and –q = [1/4πε0](qd cos θ)/r²
17. Electric field due to a dipole
Er = [1/4πε0](2p cos θ)/r³
Eθ = [1/4πε0](p sin θ)/r³
Resultant electric field at P = E= √ (Er²+ Eθ²)
= [1/4πε0](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 = ∫ 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 ε0.
In symbols
∮ E.∆s (* Intregration over a closed surface) =
qin/ε0.
Where
qin = charge enclosed by the closed surface
ε0 = emittivity of the free space
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 ε0 r²
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 ε0 R³
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 ε0r
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/2ε0
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/ ε0
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 ε0.
In symbols
∮ E.∆s (* Intregration over a closed surface) =
qin/ε0.
Where
qin = charge enclosed by the closed surface
ε0 = emittivity of the free space
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 ε0 r²
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 ε0 R³
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 ε0r
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/2ε0
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/ ε0
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.
2. For parallel plate capacitor
C = ε0A/d
A = 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 = 4πε0R1R2/[R2-R1]
If the capacitor is an isolated sphere (outer sphere is assumed to be at infinity, hence R2 is infinity and
C = 4πε0R1
V becomes Q/C = Q/4πε0R1
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 = ε0A/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 = 2πε0l/ln(R2/R1)
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 = Q²/2Aε0
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 σp 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)
= σpAl.
where
A is area of cross section of the dielectric slab
As polarization is defined as dipole moment induced per unit volume,
P = σpAl/Al (Al = volume of slab)
= σp
The induced surface charge density is equal in magnitude to the polarization P.
9. Capacitance of a parallel plate capacitor with dielectric
C = KC0
where C0 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
QP = Q[1 - (1/K)]
QP = induced charge in the dielectric
Q = Applied charge
K = dielectric constant
10. Gauss's law when dielectric materials are involved
∮KE.dS = Qfree/ε0
Where integration is over the surface, E and dS are vectors, Qfree 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 = Eε0 + P; E and P are vectors
E = electric field and P is polarisation
11. Electric field due to a point charge placed inside a dielectric
E = q/4πε0Kr²
Energy in the electric field in a dielectric
u = ½Kε0E²
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 = ε0A/d
A = 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 = 4πε0R1R2/[R2-R1]
If the capacitor is an isolated sphere (outer sphere is assumed to be at infinity, hence R2 is infinity and
C = 4πε0R1
V becomes Q/C = Q/4πε0R1
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 = ε0A/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 = 2πε0l/ln(R2/R1)
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 = Q²/2Aε0
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 σp 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)
= σpAl.
where
A is area of cross section of the dielectric slab
As polarization is defined as dipole moment induced per unit volume,
P = σpAl/Al (Al = volume of slab)
= σp
The induced surface charge density is equal in magnitude to the polarization P.
9. Capacitance of a parallel plate capacitor with dielectric
C = KC0
where C0 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
QP = Q[1 - (1/K)]
QP = induced charge in the dielectric
Q = Applied charge
K = dielectric constant
10. Gauss's law when dielectric materials are involved
∮KE.dS = Qfree/ε0
Where integration is over the surface, E and dS are vectors, Qfree 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 = Eε0 + P; E and P are vectors
E = electric field and P is polarisation
11. Electric field due to a point charge placed inside a dielectric
E = q/4πε0Kr²
Energy in the electric field in a dielectric
u = ½Kε0E²
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
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
vd = l/ τ = ½ (eE/m)τ = kE
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 = nevd
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(T0)[1 + α(T - T0)]
α is called temperature coefficient of resistivity.
6. The work done by the battery force per unit charge is
Є = W/q = Fb*d/q
This Є is called the emf of the battery. Please note that emf is work done/charge.
If nothing is externally connected
Fb = qE or
Fb*d = qEd = qV (because V =Ed)
V = potential difference between the terminals
As Є = W/q = Fb*d/q = qV/q = V
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
vd = l/ τ = ½ (eE/m)τ = kE
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 = nevd
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(T0)[1 + α(T - T0)]
α is called temperature coefficient of resistivity.
6. The work done by the battery force per unit charge is
Є = W/q = Fb*d/q
This Є is called the emf of the battery. Please note that emf is work done/charge.
If nothing is externally connected
Fb = qE or
Fb*d = qEd = qV (because V =Ed)
V = potential difference between the terminals
As Є = W/q = Fb*d/q = qV/q = V
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.
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 θc, θn, θi are temperature of the cold junction, neutral temperature and inversion temperature respectively
c
θn – θc = θi - θn
3. Thermo emf depends on temperature with the relation
ЄAB = aABθ + ½ bsub>ABθ²
Where aAB and bAB are constants for a pair of metals A and B.
This gives dЄAB/dθ = aAB + bsub>ABθ
The quantity dЄAB/dθ is called thermoelectric power at temperature θ.
5. Law of intermediate Metal
Suppose ЄAB, ЄAC, ЄBC are emfs from thermocoupes AB,AC, and BC.
If hot junctions and cold junctions of the three thermocouples are at the same temperature,
Then,
ЄAB = ЄAC - ЄBC
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. Then
Є 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 = ЄAB =
(ΠAB)T - (Π AB ) T0 +(T – T0) (σA - σA)
(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 θc, θn, θi are temperature of the cold junction, neutral temperature and inversion temperature respectively
c
θn – θc = θi - θn
3. Thermo emf depends on temperature with the relation
ЄAB = aABθ + ½ bsub>ABθ²
Where aAB and bAB are constants for a pair of metals A and B.
This gives dЄAB/dθ = aAB + bsub>ABθ
The quantity dЄAB/dθ is called thermoelectric power at temperature θ.
5. Law of intermediate Metal
Suppose ЄAB, ЄAC, ЄBC are emfs from thermocoupes AB,AC, and BC.
If hot junctions and cold junctions of the three thermocouples are at the same temperature,
Then,
ЄAB = ЄAC - ЄBC
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. Then
Є 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 = ЄAB =
(ΠAB)T - (Π AB ) T0 +(T – T0) (σA - σA)
(More explanation is needed for the formula.)
IIT JEE Physics Formula Revision 34. Magnetic Field
1. Magnetic field can be defined mathematically as
F = qv × 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 = il×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 = iA× B
iA 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 = niA× B
μ = niA
F = qv × 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 = il×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 = iA× B
iA 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 = niA× B
μ = niA
IIT JEE Physics Formula Revision 35. Magnetic field due to a Current
Note: vectors are shown in bold letters
1. Biot Savart Law
dB = [1/4 π ε0c²]*(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/ ε0c²] is written as µ0 and is called the permeability of vacuum.
Its value is 4 π*10-7 T-m/A
In terms of µ0 equation (1) becomes
dB = (µ0 /4 π)*(i) (dl*r/r³) ... (2)
The magnitude of the field
dB = (µ0 /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 = (µ0i /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 = (µ0i /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 = (µ0i /2 πd) … (6)
7. Force between two parallel currents
dF/dl = (µ0i1i2 )/2 πd
dF/dl = force per unit length of the wire W2 due to wire W1
i1,i2 = current in wire W1 and W2 respectively
d = distance between wires kept parallel to each other.
8. Magnetic field in an axial point
B = (µ0ia²)/[2(a²+d²)3/2] .. (8)
If the point is far away from the centre d is very large compared to a
B = (µ0ia²)/2d³
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
B = (µ0i /4π)(2µ/d³) .. (9)
10. Ampere’s law
The circulation of ∫B.dl of the resultant magnetic field along a closed plane curve is equal to µ0 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 = µ0*I … (10)
11. Magnetic field inside a solenoid
B = µ0ni
n = number of turns per unit length along the length of the solenoid.
i = current passing through the solenoid
12. toroid
B = µ0Ni/2πr
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
dB = [1/4 π ε0c²]*(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/ ε0c²] is written as µ0 and is called the permeability of vacuum.
Its value is 4 π*10-7 T-m/A
In terms of µ0 equation (1) becomes
dB = (µ0 /4 π)*(i) (dl*r/r³) ... (2)
The magnitude of the field
dB = (µ0 /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 = (µ0i /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 = (µ0i /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 = (µ0i /2 πd) … (6)
7. Force between two parallel currents
dF/dl = (µ0i1i2 )/2 πd
dF/dl = force per unit length of the wire W2 due to wire W1
i1,i2 = current in wire W1 and W2 respectively
d = distance between wires kept parallel to each other.
8. Magnetic field in an axial point
B = (µ0ia²)/[2(a²+d²)3/2] .. (8)
If the point is far away from the centre d is very large compared to a
B = (µ0ia²)/2d³
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
B = (µ0i /4π)(2µ/d³) .. (9)
10. Ampere’s law
The circulation of ∫B.dl of the resultant magnetic field along a closed plane curve is equal to µ0 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 = µ0*I … (10)
11. Magnetic field inside a solenoid
B = µ0ni
n = number of turns per unit length along the length of the solenoid.
i = current passing through the solenoid
12. toroid
B = µ0Ni/2πr
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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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 = (µ0/4 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 = (µ0/4 π)[(2Md)/(d² – l²)²] …(6)
If d is very large compared to l, then
B = (µ0/4 π)(2M/d³) … (7)
8. Broad-on Position
B = (µ0/4 π)[m 2l/(d² + l²)3/2 ] …(8)
= (µ0/4 π)[M/(d² + l²)3/2 ]
If d is very large compared to l,
B = (µ0/4 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
Bl = -dV/dl ... (11)
12.For a pole of pole strength m, the field at a distance r is
B = (µ0/4 π)[m/r²)
So the potential at a distance r is
V( r )= - ∞∫r (µ0/4 π)[m/r²)dr
= (µ0/4 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 = (µ0/4 π)[Mcos θ /r²)
Where
M = 2ml =magnetic moment of the dipole
14. Magnetic field due to dipole
Magnetic field at P =
(µ0/4 π)[M /r²)√(1 +3 cos² θ)] .. (14)
17. current in galvanometer
i = K tan θ .. (17)
where K = 2rBH/µ0 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
ig = [Rs/(Rs+g)]*i
i = main current
ig = current that goes through galvanometer
20. tangent law of perpendicular fields
B = BHtan θ .. (20)
21. Deflection magnetometer
Tan A position
M/ BH = (4 π/µ0)[(d² – l²)²/2d] tan θ
22. Deflection magnetometer Tan B position
M/ BH = (4 π/µ0)[(d² – l²)3/2] tan θ.. (22)
23. Oscillation Magnetometer
Time period T = 2 π/ ω = 2 π√ (I/M BH) .. (23)
From equation (23)
M BH) = 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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A magnetic charge m placed in a magnetic field B experiences a force.
F = mB .. (1)
2. Magnetic field due to magnetic charge
B = (µ0/4 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 = (µ0/4 π)[(2Md)/(d² – l²)²] …(6)
If d is very large compared to l, then
B = (µ0/4 π)(2M/d³) … (7)
8. Broad-on Position
B = (µ0/4 π)[m 2l/(d² + l²)3/2 ] …(8)
= (µ0/4 π)[M/(d² + l²)3/2 ]
If d is very large compared to l,
B = (µ0/4 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
Bl = -dV/dl ... (11)
12.For a pole of pole strength m, the field at a distance r is
B = (µ0/4 π)[m/r²)
So the potential at a distance r is
V( r )= - ∞∫r (µ0/4 π)[m/r²)dr
= (µ0/4 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 = (µ0/4 π)[Mcos θ /r²)
Where
M = 2ml =magnetic moment of the dipole
14. Magnetic field due to dipole
Magnetic field at P =
(µ0/4 π)[M /r²)√(1 +3 cos² θ)] .. (14)
17. current in galvanometer
i = K tan θ .. (17)
where K = 2rBH/µ0 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
ig = [Rs/(Rs+g)]*i
i = main current
ig = current that goes through galvanometer
20. tangent law of perpendicular fields
B = BHtan θ .. (20)
21. Deflection magnetometer
Tan A position
M/ BH = (4 π/µ0)[(d² – l²)²/2d] tan θ
22. Deflection magnetometer Tan B position
M/ BH = (4 π/µ0)[(d² – l²)3/2] tan θ.. (22)
23. Oscillation Magnetometer
Time period T = 2 π/ ω = 2 π√ (I/M BH) .. (23)
From equation (23)
M BH) = 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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