CBSE Class 12 Physics 2025: question 24

Hints

Open one hint at a time and try again before the next.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

1. (a) While the current grows, a back emf $\epsilon =-L\frac{dI}{dt}$ opposes it. The source must do work against it at the rate $\frac{dW}{dt}=|\epsilon |I=LI\frac{dI}{dt}$.

   ½ mark

2. So $dW=LIdI$.

   ½ mark

3. Total work to raise the current from 0 to I: $W={\int }_0^{I}LIdI=\frac{1}{2}L{I}^2$. This work is stored as magnetic energy.

   ½ mark

4. (b) For a long air-filled solenoid of length $l$, cross-section A and n turns per unit length: $L={\mu }_0{n}^{2}Al$ and $B={\mu }_{0}nI$, so $I=\frac{B}{{\mu }_{0}n}$.

   ½ mark

5. $U_B=\frac{1}{2}L{I}^2=\frac{1}{2}{\mu }_0{n}^{2}Al{\left(\frac{B}{{\mu }_{0}n}\right)}^2=\frac{B^{2}Al}{2{\mu }_0}$.

   ½ mark

6. Energy density $u_B=\frac{U_B}{\text{volume}}=\frac{U_B}{Al}=\frac{B^2}{2{\mu }_0}$.

   ½ mark

Answer: (a) W = ½LI²; (b) u_B = B²/(2μ₀)

Revise the chapter

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