CBSE Class 12 Physics 2026: question 33(a)

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1. (a) Faraday's law. The magnitude of the emf induced in a circuit equals the time rate of change of magnetic flux linked with it. Its direction is such that it opposes that change (Lenz's law).
   $$\epsilon =-N\frac{d{\Phi }_B}{dt}$$

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2. (b) Field and flux linkage. Let a current $I$ flow in the solenoid. The number of turns per unit length is $n=N/l$.

   For a long solenoid, the field inside is uniform:
   $$B={\mu }_{0}nI=\frac{{\mu }_{0}NI}{l}$$

   The flux through each turn is $\Phi =BA$. So the total flux linkage is:
   $$N\Phi =N\cdot \frac{{\mu }_{0}NI}{l}\cdot A=\frac{{\mu }_0{N}^{2}A}{l}I$$

   1 mark

3. Self-inductance. Since $N\Phi =LI$:
   $$\boxed{L=\frac{{\mu }_0{N}^{2}A}{l}}(\text{or }{\mu }_0{n}^{2}Al)$$

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4. (c) Convert the data to SI units.

   • $l=50$ cm $=0.5$ m
   • $B=4.0$ mT $=4.0\times 10^{-3}$ T
   • $\omega =60$ rpm $=\frac{60\times 2\pi }{60}=2\pi$ rad s${}^{-1}$

   For a rod rotating about one end, the emf is:
   $$\epsilon =\frac{1}{2}B\omega l^2$$

   1 mark

5. Substitute.
   $$\epsilon =\frac{1}{2}\times 4.0\times 10^{-3}\times 2\pi \times (0.5{)}^2=\pi \times 10^{-3}\text{ V}$$
   $$\approx 3.14\times 10^{-3}\text{ V}=3.14\text{ mV}$$

   1 mark

Answer: (a) $\epsilon =-Nd\Phi /dt$; (b) $L={\mu }_0{N}^{2}A/l$; (c) $\epsilon \approx 3.14\times 10^{-3}$ V

Revise the chapter

• Chapter: Electromagnetic Induction (notes, formulas, where marks are lost and every past question)
• Topic: Self and Mutual Inductance