JEE Physics Electromagnetic Induction Notes Set 03

Magnetic Flux, Faraday's Law and Lenz's Law

The flux through a surface is \( \Phi=\int\vec B\cdot d\vec S \), measured in weber. A uniform field gives \( \Phi=BA\cos\theta \), where \( \theta \) is the angle between B and the normal. An emf appears in a circuit whenever the flux linked with it changes, for any reason: B changes, the area changes, or the orientation changes.
Faraday's law: \( \varepsilon=-N\frac{d\Phi}{dt} \). The magnitude is the rate of change of flux linkage; the minus sign is Lenz's law, which says the induced current always opposes the change that causes it. In practice find the magnitude from the rate, then find the direction by asking which way the induced field must point to oppose the change.
Charge that flows in a closed circuit of resistance R is \( q=\frac{|\Delta\Phi|}{R} \) (times N for N turns), independent of how fast the change occurs, while the heat produced depends on the whole time history: \( H=\int \frac{\varepsilon^2}{R}dt \). Lenz's law is energy conservation: a magnet approaching a ring is repelled, a receding one is attracted, and the work done against that force appears as heat. A ring with a cut carries no current, so it exerts no braking force.

Motional EMF

A straight conductor of length \( \ell \) moving with velocity v in a uniform field B develops \( \varepsilon=(\vec v\times\vec B)\cdot\vec\ell \), which is \( B\ell v \) when v, B and the rod are mutually perpendicular. The magnetic force pushes free charges along the rod until the electric field they set up satisfies \( E=vB \). The end reached by positive charge is at higher potential, so the direction follows \( \vec v\times\vec B \).
For a bent wire in a uniform field the emf between two points depends only on the straight line joining them (effective length). A closed loop moving in a uniform field has zero net emf. If B varies along the rod, integrate: \( \varepsilon=\int vB(x)\,dx \).
A rod sliding on rails with a resistor R behaves like a battery of emf \( B\ell v \) and gives current \( i=\frac{B\ell v}{R} \), a retarding force \( \frac{B^2\ell^2v}{R} \), and power \( \frac{B^2\ell^2v^2}{R} \) equal to the rate of work of the agent pulling it at constant speed. Under a constant force F the speed rises to the terminal value \( v_T=\frac{FR}{B^2\ell^2} \) exponentially. A capacitor across the rails behaves like an extra mass \( CB^2\ell^2 \) and a falling rod then has constant acceleration \( \frac{mg}{m+CB^2\ell^2} \).

Rotation, Discs and Non-uniform Fields

A rod of length \( \ell \) rotating with angular speed \( \omega \) about one end in a field perpendicular to its plane has emf \( \frac12B\omega\ell^2 \); if the pivot lies inside the rod, subtract the two end contributions: \( \frac12B\omega(r_2^2-r_1^2) \). A rod pivoted at its midpoint has zero emf between its ends, though each half has emf. A conducting disc of radius R rotating in a perpendicular field gives \( \frac12B\omega R^2 \) between centre and rim.
A coil of N turns and area A rotating at \( \omega \) in field B has \( \varepsilon=NBA\omega\sin\omega t \): the peak emf occurs when the plane is parallel to B (flux zero) and the emf is zero when the flux is maximum. The average over a half-turn is \( \frac2\pi \) of the peak and over a full turn is zero.
Near a long straight wire carrying i, the field falls as \( \frac{\mu_0i}{2\pi r} \). The flux through a rectangle of sides a (parallel to the wire) and b, with near edge at distance d, is \( \frac{\mu_0ia}{2\pi}\ln\frac{d+b}{d} \). For a loop moving away, the emf comes from the two sides parallel to the wire and the nearer side dominates.

Induced Electric Field of a Changing Magnetic Field

A changing B creates a non-conservative electric field whose line integral around a loop equals the emf: \( \oint\vec E\cdot d\vec l=-\frac{d\Phi}{dt} \). For a uniform field changing at rate \( \frac{dB}{dt} \) in a cylinder of radius R, the field lines are circles: inside, \( E=\frac r2\frac{dB}{dt} \); outside, \( E=\frac{R^2}{2r}\frac{dB}{dt} \). The field exists even where B itself is zero.
Because the field is tangential, radial wire segments have no emf in it. The emf in a straight chord is therefore the rate of change of B times the area of the triangle formed by the chord and the two radii to its ends. Electrons or charges released in such a region are accelerated tangentially.

Self-Inductance and Magnetic Energy

The flux linked with a coil is proportional to its current: \( N\Phi=Li \), and the self-induced emf is \( -L\frac{di}{dt} \). A long solenoid of N turns, length \( \ell \) and area A has \( L=\frac{\mu_0N^2A}{\ell}=\mu_0n^2A\ell \). With a fixed length of wire, a solenoid of smaller radius has more turns, so compare shapes through \( \frac{N^2A}{\ell} \).
An inductor stores energy \( U=\frac12Li^2 \) with energy density \( u=\frac{B^2}{2\mu_0} \) in the field. The rate of storage is \( Li\frac{di}{dt} \). When the current in an inductor increases, its end at which the current enters is at the higher potential, with voltage \( L\frac{di}{dt} \); when it decreases the polarity reverses.
The current through an inductor cannot change instantly: at \( t=0 \) it acts as an open circuit, and at steady state it acts as a plain wire (apart from its own resistance). If L changes abruptly, it is the flux linkage \( Li \) that stays constant.

LR Circuits

Growth with battery E: \( i=\frac ER\left(1-e^{-t/\tau}\right) \), \( \tau=\frac LR \). The inductor voltage is \( Ee^{-t/\tau} \), the resistor voltage is \( E(1-e^{-t/\tau}) \), and the two always add to E. At \( t=\tau \) the current is 63 percent of its final value; half the final value is reached at \( t=\tau\ln2 \).
Decay (battery removed, loop closed): \( i=i_0e^{-t/\tau} \), and the whole stored energy \( \frac12Li_0^2 \) is dissipated as heat in the resistance. The charge passed during growth up to time t is \( \frac ER\left[t-\tau(1-e^{-t/\tau})\right] \).
For two inductors in series with no mutual coupling \( L=L_1+L_2 \); in parallel \( \frac1L=\frac1{L_1}+\frac1{L_2} \) and the currents divide in inverse ratio of inductance at every instant. In the first instant the current in an LR branch is blocked; later the inductor's resistance alone limits it.

Mutual Inductance

The flux linked with coil 2 due to the current in coil 1 is \( N_2\Phi_{21}=M\,i_1 \), and \( M_{12}=M_{21} \). The emf in either coil is \( M\frac{di}{dt} \) from the other. For two coaxial solenoids, one inside the other, \( M=\mu_0n_1n_2\pi r_1^2\ell \) with r the radius of the inner one. A straight wire along the axis of a ring has \( M=0 \) because its field lies in the plane of the ring.
Always compute whichever flux is easier: for a very small loop at the centre of a big one, find the big loop's field at the centre and multiply by the small area. The coupling coefficient is \( k=\frac{M}{\sqrt{L_1L_2}}\le1 \). Two coils in series have \( L=L_1+L_2\pm2M \), with the plus sign when their fluxes add.

LC Oscillations

A charged capacitor connected to an inductor with negligible resistance oscillates: \( q=q_0\cos\omega t \), \( i=-q_0\omega\sin\omega t \), \( \omega=\frac1{\sqrt{LC}} \), \( T=2\pi\sqrt{LC} \). The total energy \( \frac{q_0^2}{2C} \) stays constant and moves between electric and magnetic forms, the energy being all electric at \( t=0,\frac T2 \), all magnetic at \( \frac T4,\frac{3T}4 \), and shared equally at \( \frac T8,\frac{3T}8,\ldots \).
The circuit is the electrical counterpart of a mass on a spring: L plays the role of mass, 1/C of the spring constant and q of displacement. Series inductors add; parallel capacitors add. Tuning a radio means varying C to change the resonant frequency \( f=\frac1{2\pi\sqrt{LC}} \).

Worked Examples

Worked Example 1. A coil of 100 turns and area 0.02 m^2 is placed with its plane perpendicular to a field that falls uniformly from 0.5 T to 0.1 T in 0.2 s. The resistance of the coil is 10 ohm. Find the induced emf, the current and the charge that flows.
Answer:
\( \varepsilon=N A\frac{\Delta B}{\Delta t}=100\times0.02\times\frac{0.4}{0.2}=4 \) V.
\( i=\frac{4}{10}=0.4 \) A.
\( q=i\,\Delta t=0.4\times0.2=0.08 \) C (also \( \frac{N A\Delta B}{R}=\frac{100\times0.02\times0.4}{10}=0.08 \) C).

Worked Example 2. A rod of length 0.5 m and resistance 2 ohm slides at 4 m/s on smooth rails (resistance zero) joined by a 3 ohm resistor, in a field of 0.25 T perpendicular to the plane. Find the current, the force needed to hold the speed constant, and the power.
Answer:
Emf: \( B\ell v=0.25\times0.5\times4=0.5 \) V.
Total resistance 5 ohm, so \( i=0.1 \) A.
Force: \( F=Bi\ell=0.25\times0.1\times0.5=0.0125 \) N, and power \( Fv=0.05 \) W, which equals \( i^2R=0.01\times5 \) W.

Worked Example 3. A metal rod of length 1.2 m rotates at 20 rad/s about an axis through one end, perpendicular to the rod, in a uniform field of 0.05 T parallel to the axis. Find the emf between its ends.
Answer:
\( \varepsilon=\frac12B\omega\ell^2=\frac12\times0.05\times20\times1.44 \)
\( \varepsilon=0.72 \) V.

Worked Example 4. A uniform field in a cylinder of radius 5 cm increases at 0.2 T/s. Find the induced electric field at 3 cm and at 10 cm from the axis.
Answer:
Inside: \( E=\frac r2\frac{dB}{dt}=\frac{0.03}{2}\times0.2=3\times10^{-3} \) V/m.
Outside: \( E=\frac{R^2}{2r}\frac{dB}{dt}=\frac{0.0025}{0.2}\times0.2=2.5\times10^{-3} \) V/m.

Worked Example 5. An inductor of 0.5 H and resistance 5 ohm is joined to a 10 V battery. Find the time for the current to reach 1 A, the final current and the energy stored finally.
Answer:
Final current \( i_0=\frac{10}{5}=2 \) A; \( \tau=\frac{0.5}{5}=0.1 \) s.
\( 1=2(1-e^{-t/0.1}) \Rightarrow t=0.1\ln2=0.0693 \) s.
\( U=\frac12\times0.5\times4=1 \) J.

Worked Example 6. An LC circuit has L = 20 mH and C = 50 microfarad, with the capacitor initially charged to 100 V. Find the angular frequency, the peak current and the time at which the energy is first shared equally.
Answer:
\( \omega=\frac1{\sqrt{LC}}=\frac{1}{\sqrt{2\times10^{-2}\times5\times10^{-5}}}=\frac1{10^{-3}}=1000 \) rad/s.
Peak current: \( q_0\omega=CV\omega=5\times10^{-5}\times100\times1000=5 \) A.
\( t=\frac T8=\frac{\pi}{4\omega}=7.85\times10^{-4} \) s.