CBSE · Class 12 · Physics · Electromagnetic InductionAn inductor of self-inductance $2.0\text{ H}$ carries a current of $10\text{ A}$. If the current drops uniformly to zero in $0.02\text{ s}$, calculate: (i) The average self-induced electromotive force (emf) developed across the inductor. (ii) The magnetic potential energy stored in the inductor initially.

Step-by-Step Solution:

1. Given Data:

• Self-inductance, $L = 2.0\text{ H}$
• Initial current, $I_1 = 10\text{ A}$
• Final current, $I_2 = 0\text{ A}$
• Time interval, $\Delta t = 0.02\text{ s}$

(i) Calculation of Average Induced EMF ($e$):\nChange in current, $\Delta I = I_2 - I_1 = 0 - 10 = -10\text{ A}$ \nRate of change of current, $\frac{\Delta I}{\Delta t} = \frac{-10\text{ A}}{0.02\text{ s}} = -500\text{ A/s}$ \nUsing the formula for self-induced emf: $$e = -L \frac{\Delta I}{\Delta t}$$ $$e = -(2.0) \times (-500\text{ A/s}) = 1000\text{ V}$$

(ii) Calculation of Initial Stored Magnetic Energy ($U$):\nThe energy stored in an inductor carrying current $I_1$ is given by: $$U = \frac{1}{2} L I_1^2$$ $$U = \frac{1}{2} \times 2.0 \times (10)^2$$ $$U = 1.0 \times 100 = 100\text{ J}$$

Final Answer: (i) The average self-induced emf is $1000\text{ V}$ (or $1\text{ kV}$). (ii) The initial magnetic energy stored is $100\text{ J}$.