Q39

5 marksLong AnswerSection E

(a) State Faraday's law of electromagnetic induction.

(b) Derive an expression for the self-inductance of an air-filled long solenoid of length l and cross-sectional area A having N turns.

(c) A conducting rod of length 50 cm, with one end pivoted, is rotated with angular speed of 60 rpm in a uniform magnetic field of 4.0 mT directed perpendicular to the plane of rotation of rod. Find the emf induced in the rod.

OR

(a) Draw a labelled diagram of a step-up transformer. State the principle on which it works and obtain the ratio of secondary voltage to primary voltage in terms of number of turns and currents in the two coils.

(b) The ratio of the number of turns in the primary to the secondary of an ideal transformer is $1 : 5$ . If 5 kW power at 200 V is supplied to the primary, find (i) current in the primary, and (ii) output voltage.

Q: Why does the self-inductance of a solenoid depend on $N^2$?

The magnetic field inside is $B = \mu_0(N/l)I$, so it is proportional to N. The flux through one turn is $B\cdot A$, and the total flux linkage is that flux multiplied by the N turns — giving a second factor of N. Hence total linkage $\propto N^2$, and since L = (total flux linkage)/I, we get $L = \mu_0 N^2 A/l$.

Electromagnetic Induction

Faraday's Law, Self-Inductance of a Solenoid and Motional EMF

Official Answer

This 5-mark answer states Faraday's law, derives the solenoid self-inductance, and computes the emf of a rotating rod.

Part (a) — Faraday's law

First law: Whenever the magnetic flux linked with a circuit changes, an emf is induced; it lasts only while the flux is changing.
Second law: The magnitude of the induced emf equals the rate of change of magnetic flux linkage:

$\varepsilon = -\, d\Phi_B/dt$ (for N turns, $\varepsilon = -N\, d\Phi_B/dt$ ). The minus sign is Lenz's law.

Part (b) — Self-inductance of a long solenoid

Number of turns per unit length: $n = N/l$ .
Magnetic field inside: $B = \mu_0 n I = \mu_0 (N/l) I$ .
Flux through one turn: $\Phi = B\cdot A = \mu_0 (N/l) I A$ .
Total flux linkage: $N\Phi = \mu_0 (N^2/l) A I$ .
Since $N\Phi = L I$ :

$L = \mu_0 N^2 A / l$ .

Part (c) — EMF of the rotating rod

Given $L = 0.50\text{ m}$ , $B = 4.0 \times 10^{-3}\text{ T}$ , rotation = $60\text{ rpm} = 1\text{ rev/s} \to \omega = 2\pi\text{ rad/s}$ .

Motional emf of a rod rotating about one end: $\varepsilon = \frac{1}{2} B \omega L^2$

$\varepsilon = \frac{1}{2} \times (4.0 \times 10^{-3}) \times (2\pi) \times (0.50)^2 = \frac{1}{2} \times 4.0 \times 10^{-3} \times 2\pi \times 0.25$

$\varepsilon = \pi \times 10^{-3}\text{ V} \approx 3.14 \times 10^{-3}\text{ V} \approx 3.14\text{ mV}$ .

OR alternative (transformer): For an ideal transformer, $V_s/V_p = N_s/N_p = I_p/I_s$ . With $N_p:N_s = 1:5$ , 5 kW at 200 V: primary current $I_p = P/V_p = 5000/200$ = 25 A; output voltage $V_s = V_p \times (N_s/N_p) = 200 \times 5$ = 1000 V.

Faraday's lawflux linkageself-inductancesolenoidmotional emfrotating rodangular speedLenz's law

Marking Scheme

11 mark: statement of Faraday's law (induced emf = rate of change of flux linkage, $\varepsilon = -N\, d\Phi/dt$ ).
21.5 marks: solenoid derivation — $B = \mu_0 nI$ , flux linkage $N\Phi$ , $L = \mu_0 N^2 A/l$ .
30.5 mark: correct final self-inductance expression $L = \mu_0 N^2 A/l$ .
41 mark: convert 60 rpm to $\omega = 2\pi\text{ rad/s}$ and use $\varepsilon = \frac{1}{2}B\omega L^2$ .
51 mark: $\varepsilon = \pi \times 10^{-3}\text{ V} \approx 3.14\text{ mV}$ with SI units.
6OR: (a) 2 marks labelled diagram + principle (mutual induction) + $V_s/V_p = N_s/N_p = I_p/I_s$ ; (b) 1.5 marks $I_p = 25\text{ A}$ , 1.5 marks $V_s = 1000\text{ V}$ .

Hint

Faraday: $\varepsilon = -N\, d\Phi/dt$ . Solenoid: $B = \mu_0 nI, L = N\Phi/I \to \mu_0 N^2 A/l$ . Rod: convert 60 rpm to $\omega = 2\pi\text{ rad/s}$ , then $\varepsilon = \frac{1}{2}B\omega L^2$ .

Quick Oral Answer

Faraday's law: induced emf equals the rate of change of flux linkage, $\varepsilon = -N\, d\Phi/dt$ . A long solenoid has $L = \mu_0 N^2 A/l$ . A rod of length 0.5 m spinning at 60 rpm ( $\omega = 2\pi\text{ rad/s}$ ) in a 4 mT field gives $\varepsilon = \frac{1}{2}B\omega L^2 \approx 3.14\text{ mV}$ .

Analysis & Explanation

Concept. This question threads together the three pillars of electromagnetic induction: Faraday's flux rule, the geometric derivation of a solenoid's self-inductance, and the motional emf of a rod sweeping through a field.

Self-inductance intuition. $L = \mu_0 N^2 A/l$ depends on geometry alone (turns, area, length) and on the core material — not on the current. The N² dependence is why tightly wound multi-turn coils store large magnetic energy, the principle behind inductors and chokes.

The rotating-rod result. Each element of the rod moves at a different speed, so you integrate: $\varepsilon = \int_0^L B(\omega x)dx = \frac{1}{2}B\omega L^2$ . Equivalently, the rod sweeps area $\frac{1}{2}L^2$ per radian, giving flux rate ½BωL². A frequent slip is using $\varepsilon = B\omega L^2$ (missing the factor ½) or forgetting to convert rpm to rad/s.

Numerical care. 60 rpm = 1 revolution per second, so $\omega = 2\pi\text{ rad/s}$ — not 60 rad/s. Keeping SI units (B in tesla, L in metres) gives $\varepsilon = \pi\text{ mV}$ .

Real-world link. Rotating-conductor emf is the basis of the AC generator; self-inductance governs how coils oppose current changes in every power supply; and the transformer of the OR-part — stepping 200 V up to 1000 V — is exactly how the grid transmits power efficiently at high voltage and low current.

OR part. The transformer relation $V_s/V_p = N_s/N_p = I_p/I_s$ follows from equal flux per turn and (ideally) conserved power; a step-up in voltage is a step-down in current.

Common Mistakes

1Using $\varepsilon = B\omega L^2$ without the factor $\frac{1}{2}$ — the rod rotating about one end sweeps area $\frac{1}{2}L^2$ per radian, so $\varepsilon = \frac{1}{2}B\omega L^2$ .
2Forgetting to convert 60 rpm to $\omega = 2\pi\text{ rad/s}$ (using 60 rad/s gives an answer ~9.5× too large).
3Writing self-inductance as $\mu_0 NA/l$ instead of $\mu_0 N^2 A/l$ — dropping one factor of N from the flux linkage.

Interesting Facts

The self-inductance $L = \mu_0 N^2 A/l$ depends only on the coil's geometry and core, never on the current flowing — doubling the turns quadruples the inductance.

A rod pivoted at one end and spun in Earth's field generates a real (if tiny) emf — the same 'homopolar generator' idea Michael Faraday demonstrated in 1831.

Step-up transformers raise generated voltages to hundreds of kilovolts for transmission, cutting $I^2R$ line losses — the reason AC beat DC in the 1890s 'War of Currents'.

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Frequently Asked Questions

Why does the self-inductance of a solenoid depend on $N^2$ ?

The magnetic field inside is $B = \mu_0(N/l)I$ , so it is proportional to N. The flux through one turn is $B\cdot A$ , and the total flux linkage is that flux multiplied by the N turns — giving a second factor of N. Hence total linkage $\propto N^2$ , and since L = (total flux linkage)/I, we get $L = \mu_0 N^2 A/l$ .

Why is the emf of the rotating rod $\frac{1}{2}B\omega L^2$ and not BωL²?

Different points on the rod move at different linear speeds ( $v = \omega x$ increases with distance x from the pivot). Integrating the motional emf $B\cdot v$ across the rod, $\varepsilon = \int_0^L B\omega x\, dx = \frac{1}{2}B\omega L^2$ . The factor ½ comes from averaging the linear speed along the length; equivalently the rod sweeps an area $\frac{1}{2}L^2$ per radian.

How do you get ω from 60 rpm?

60 revolutions per minute = $60/60 = 1$ revolution per second. Each revolution is 2π radians, so $\omega = 2\pi \times 1 = 2\pi\text{ rad/s} \approx 6.28\text{ rad/s}$ . Using this in $\varepsilon = \frac{1}{2}B\omega L^2$ with B = 4.0 mT and L = 0.5 m gives $\varepsilon = \pi \times 10^{-3}\text{ V} \approx 3.14\text{ mV}$ .