JEE Physics Electromagnetic Induction Notes Set 01

Magnetic Flux

The flux through a surface is \( \phi=\int\vec B\cdot d\vec A=BA\cos\theta \) for a uniform field, where \( \theta \) is the angle between B and the normal to the surface. The SI unit is the weber (1 Wb = 1 T m²). Flux through a coil of N turns, counting all turns, is N times that through one turn.

Faraday's Law and Lenz's Law

Whenever the flux linked with a circuit changes, an emf is induced: \( \varepsilon=-N\frac{d\phi}{dt} \). The minus sign expresses Lenz's law: the induced current flows in the direction that opposes the change in flux that produced it. This is a statement of energy conservation, since pushing a magnet towards a closed loop needs work against the induced current.
If the circuit has resistance R, the induced current is \( I=\frac{|\varepsilon|}{R} \) and the charge that flows is \( Q=\frac{N\,\Delta\phi}{R} \), independent of the time taken.
Flux can change because B changes, the area changes or the orientation changes. A coil turned through 180 degrees from a perpendicular position changes its flux by 2NBA.

Motional emf

A straight conductor of length l moving with velocity v perpendicular to a field B has emf \( \varepsilon=Blv \). On rails with a resistor R the current is \( I=\frac{Blv}{R} \), the retarding force is \( F=\frac{B^2l^2v}{R} \) and the power supplied by the agent equals the heat produced, \( \frac{B^2l^2v^2}{R} \).
A rod of length l rotating about one end with angular speed \( \omega \) in a perpendicular field has emf \( \tfrac12B\omega l^2 \). A disc of radius R rotating about its axis gives \( \tfrac12B\omega R^2 \) between centre and rim, and zero between two rim points.
Eddy currents are loops of induced current in bulk conductors; they cause heating, which is reduced by laminating the core, and are put to use in induction furnaces, magnetic braking and energy meters.

Self Inductance

A changing current in a coil induces an emf in the same coil: \( \varepsilon=-L\frac{dI}{dt} \), where \( L=\frac{N\phi}{I} \) is the self inductance (henry). It acts as the electrical inertia of the circuit.
Long solenoid: \( L=\mu_0n^2Al=\frac{\mu_0N^2A}{l} \) (multiply by \( \mu_r \) with a core). Toroid: \( L=\frac{\mu_0N^2A}{2\pi r} \).
Energy stored in an inductor: \( U=\tfrac12LI^2 \); energy density of a magnetic field \( u=\frac{B^2}{2\mu_0} \).

LR Circuits

Growth with a battery V: \( I=\frac VR\left(1-e^{-t/\tau}\right) \) with time constant \( \tau=\frac LR \). At t = τ the current is 63% of its final value. Decay through a closed loop: \( I=I_0e^{-t/\tau} \), falling to 37% at t = τ. At the instant of closing the inductor behaves as an open circuit; in the steady state it behaves as a plain wire.
Inductors in series (no coupling) add: \( L=L_1+L_2 \); in parallel: \( \frac1L=\frac1{L_1}+\frac1{L_2} \).

Mutual Inductance, Transformer and Generator

Two coils: \( \varepsilon_2=-M\frac{dI_1}{dt} \), with \( M=\frac{N_2\phi_2}{I_1} \). For two coaxial solenoids of equal length and area: \( M=\frac{\mu_0N_1N_2A}{l} \). The coupling satisfies \( M=k\sqrt{L_1L_2} \) with \( k\le1 \).
Ideal transformer: \( \frac{V_s}{V_p}=\frac{N_s}{N_p}=\frac{I_p}{I_s} \). It works only with changing current, so a battery on the primary gives no secondary emf. Real transformers lose energy to copper heating, eddy currents, hysteresis and flux leakage; efficiency is \( \frac{P_{out}}{P_{in}} \).
AC generator: a coil of N turns rotating at angular speed \( \omega \) in a field B produces \( \varepsilon=NBA\omega\sin\omega t \), with peak value \( NBA\omega \). The emf reverses twice per revolution and is zero when the coil plane is perpendicular to the field.

Solved Examples

Illustration 1. A coil of 200 turns and area 0.02 m² is turned through 180 degrees in 0.05 s in a field of 0.5 T initially perpendicular to it. Find the average emf.
Answer:
\( \varepsilon=\frac{2NBA}{t}=\frac{2\times200\times0.5\times0.02}{0.05}=80 \) V.
Average emf = 80 V.

Illustration 2. A rod of length 0.5 m moves at 4 m/s perpendicular to a field of 0.2 T on rails joined by a 2 ohm resistor. Find the emf, the current and the force needed.
Answer:
\( \varepsilon=Blv=0.2\times0.5\times4=0.4 \) V.
\( I=\frac{0.4}{2}=0.2 \) A.
\( F=BIl=0.2\times0.2\times0.5=0.02 \) N.
0.4 V, 0.2 A, 0.02 N.

Illustration 3. An inductor of 2 H and resistance 4 ohm is connected to a 12 V battery. Find the final current, the time constant, and the current after 0.5 s.
Answer:
\( I_\infty=\frac{12}{4}=3 \) A and \( \tau=\frac24=0.5 \) s.
\( I=3(1-e^{-1})=1.90 \) A.
3 A, 0.5 s, 1.9 A.

Illustration 4. A solenoid of 1000 turns, length 0.5 m and area 4 cm squared carries 5 A. Find its self inductance and the energy stored.
Answer:
\( L=\frac{\mu_0N^2A}{l}=\frac{4\pi\times10^{-7}\times10^6\times4\times10^{-4}}{0.5}=1.0\times10^{-3} \) H.
\( U=\tfrac12LI^2=\tfrac12\times1.0\times10^{-3}\times25=1.26\times10^{-2} \) J.
L about 1 mH; U about 0.0126 J.

Illustration 5. A step-down transformer has 2000 primary turns and 100 secondary turns on a 220 V supply and delivers 5 A with 80% efficiency. Find the secondary voltage and the primary current.
Answer:
\( V_s=220\times\frac{100}{2000}=11 \) V.
Output power \( =11\times5=55 \) W, input \( =\frac{55}{0.8}=68.75 \) W, \( I_p=\frac{68.75}{220}=0.3125 \) A.
11 V and about 0.31 A.