Question 13: Electromagnetic Induction

Question

The current in a coil changes steadily from $$3 \mathrm{~A}$$ to $$5 \mathrm{~A}$$ in $$0.2 \mathrm{~s}$$ when an emf of $$2 \mu \mathrm{V}$$ is induced in it. The self-inductance of the coil is

Accepted Answer

The self-inductance of the coil can be determined using the formula for the induced emf in a coil due to a changing current. The formula is given by:
$$ \mathcal{E} = -L \frac{dI}{dt} $$
Where:
$$\mathcal{E}$$ is the induced emf
$$L$$ is the self-inductance of the coil
$$\frac{dI}{dt}$$ is the rate of change of current
We are given the following values:
$$ \mathcal{E} = 2 \mu \mathrm{V} = 2 	imes 10^{-6} \mathrm{V} $$
Initial current, $$ I_1 = 3 \mathrm{~A} $$
Final current, $$ I_2 = 5 \mathrm{~A} $$
Time interval, $$ \Delta t = 0.2 \mathrm{~s} $$
The rate of change of current can be calculated as:
$$ \frac{dI}{dt} = \frac{I_2 - I_1}{\Delta t} = \frac{5 \mathrm{~A} - 3 \mathrm{~A}}{0.2 \mathrm{~s}} = \frac{2 \mathrm{~A}}{0.2 \mathrm{~s}} = 10 \mathrm{~A/s} $$
Now, using the formula for induced emf:
$$ 2 	imes 10^{-6} \mathrm{V} = L 	imes 10 \mathrm{~A/s} $$
Solving for $$L$$:
$$ L = \frac{2 	imes 10^{-6} \mathrm{V}}{10 \mathrm{~A/s}} $$
$$ L = 2 	imes 10^{-7} \mathrm{~H} $$
$$ L = 0.2 \mu \mathrm{H} $$
Therefore, the self-inductance of the coil is 0.2 $$\mu \mathrm{H}$$. So, the correct option is:
Option D: 0.2 $$\mu H$$

Answer

0.2 mH

Answer

20 $$\mu H$$

Answer

2 $$\mu H$$

Answer

0.2 $$\mu H$$

Electromagnetic Induction

The current in a coil changes steadily from $3 \mathrm{~A} to 5 \mathrm{~A} in 0.2 \mathrm{~s} when an emf of 2 \mu \mathrm{V}$ is induced in it. The self-inductance of the coil is

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