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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Friday, February 8, 2008
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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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
I = M/V
2. Magnetic intensity
H = B/µ0 - I .. (2)
Where
H = magnetic intensity
B - resultant magnetic field
I = intensity of magnetization
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
I = χH … (6)
Χ is called the susceptibility of the material.
7. Permeability
B = µH … (7)
µ = µ0 (1+χ) is a constant and is called the permeability of the material.
µ0 is the permeability of vacuum.
µr = µ/µ0 = 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 - Tc)
Where
Tc is the Curie point and
c’ = constant
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1. Magnetization vector = Magnetic moment per unit volume
I = M/V
2. Magnetic intensity
H = B/µ0 - I .. (2)
Where
H = magnetic intensity
B - resultant magnetic field
I = intensity of magnetization
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
I = χH … (6)
Χ is called the susceptibility of the material.
7. Permeability
B = µH … (7)
µ = µ0 (1+χ) is a constant and is called the permeability of the material.
µ0 is the permeability of vacuum.
µr = µ/µ0 = 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 - Tc)
Where
Tc is the Curie point and
c’ = 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 = ∫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 = µ0n² πr² l … (7)
8. Growth of current through an LR circuit
i = i0(1 - e-tR/L) … (8)
= i0(1 - e-t/ τ ) … (9)
where
i = current in the circuit at time t
i0 = Є/R
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 = i0(1 - e-tR/L) … (10)
= i0(1 - e-t/ τ ) … (11)
i = current in the circuit at time t
i0 = current in the circuit at time t = 0
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 = B²/2µ0
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 = µ0n² πr² l … (7)
8. Growth of current through an LR circuit
i = i0(1 - e-tR/L) … (8)
= i0(1 - e-t/ τ ) … (9)
where
i = current in the circuit at time t
i0 = Є/R
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 = i0(1 - e-tR/L) … (10)
= i0(1 - e-t/ τ ) … (11)
i = current in the circuit at time t
i0 = current in the circuit at time t = 0
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 = B²/2µ0
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 = i0 sin (ωt + φ) .. (1)
where
i = current at time t
i0 = peak current
current repeats after each time interval T = 2 π/ω
2. Emf produced in a AC generator
Є = NBA ω sin ωt = Є0 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
irms = i0/√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 = i0 sin ωt …… (4)
where
i0 = Є0/R ... (5)
6. Ac circuit containing only a capacitor
i = i0 cos ωt … (6)
where
i0 =C Є0ω
= Є0/(1/ωC) … (7)
8. AC circuit containing only an inductor
i = (Є0/ωL) sin (ωt – π/2) …(8)
or i = i0 sin (ωt – π/2)
where i0 = Є0/ωL
10. Impedance
The peak current and the peak emf in Ac circuits may be written as
i0 = Є0/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 = (Є0/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 = Єrms irmscos φ
16. Voltages in primary and secondary of transformer
Є2 = - (N2 / N1) Є1 .... (16)
Power transfer
Є1i1 = Є2 i2 …(17)
i2 = - (N1 / N2) i1
‘-‘ sign shows that i2 is 180° out of phase with i1 .
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1. Alternating current
i = i0 sin (ωt + φ) .. (1)
where
i = current at time t
i0 = peak current
current repeats after each time interval T = 2 π/ω
2. Emf produced in a AC generator
Є = NBA ω sin ωt = Є0 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
irms = i0/√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 = i0 sin ωt …… (4)
where
i0 = Є0/R ... (5)
6. Ac circuit containing only a capacitor
i = i0 cos ωt … (6)
where
i0 =C Є0ω
= Є0/(1/ωC) … (7)
8. AC circuit containing only an inductor
i = (Є0/ωL) sin (ωt – π/2) …(8)
or i = i0 sin (ωt – π/2)
where i0 = Є0/ωL
10. Impedance
The peak current and the peak emf in Ac circuits may be written as
i0 = Є0/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 = (Є0/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 = Єrms irmscos φ
16. Voltages in primary and secondary of transformer
Є2 = - (N2 / N1) Є1 .... (16)
Power transfer
Є1i1 = Є2 i2 …(17)
i2 = - (N1 / N2) i1
‘-‘ sign shows that i2 is 180° out of phase with i1 .
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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 = E0 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 = B0 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 = µ0(i + id)
id = ε0*(d ΦE/dt)
Where
ΦE/ = 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/ ε0
Gauss’s law for magnetism
∮B.ds = 0
Faraday’s law
∮E.dl = -dΦB/dt
Ampere’s law
∮B.dl = µ0(i + id)
id = ε0*(d ΦE/dt)
These equations are satisfied by a plane electromagnetic wave given by
Ey = E = E0 sin ω(t-x/c)
Bz = B = B0 sin ω(t-x/c)
4. c = i/√(µ0ε0)
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 ε0E²dV + B²dV/2µ0
When we substitute the values of E and B in the above equation and take an average over a longer period of time
uav = ½ ε0E0² = B0²/2µ0
6. intensity of electromagnetic wave is per unit area per unit time I = U/AΔt = uavc.
In terms of maximum electric field (substituting the value of uavc)
I = ½ ε0E0²c
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 = E0 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 = B0 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 = µ0(i + id)
id = ε0*(d ΦE/dt)
Where
ΦE/ = 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/ ε0
Gauss’s law for magnetism
∮B.ds = 0
Faraday’s law
∮E.dl = -dΦB/dt
Ampere’s law
∮B.dl = µ0(i + id)
id = ε0*(d ΦE/dt)
These equations are satisfied by a plane electromagnetic wave given by
Ey = E = E0 sin ω(t-x/c)
Bz = B = B0 sin ω(t-x/c)
4. c = i/√(µ0ε0)
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 ε0E²dV + B²dV/2µ0
When we substitute the values of E and B in the above equation and take an average over a longer period of time
uav = ½ ε0E0² = B0²/2µ0
6. intensity of electromagnetic wave is per unit area per unit time I = U/AΔt = uavc.
In terms of maximum electric field (substituting the value of uavc)
I = ½ ε0E0²c
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.
Thursday, February 7, 2008
IIT JEE Physics Formula Revision 41. Electric Current through Gases
1. 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
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
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