Question 65: Electromagnetic Induction

Question

In a coil, the current changes from $$-2 \mathrm{~A}$$ to $$+2 \mathrm{~A}$$ in $$0.2 \mathrm{~s}$$ and induces an emf of $$0.1 \mathrm{~V}$$. The self inductance of the coil is :

Accepted Answer

To find the self-inductance of the coil, we can use the formula for induced electromotive force (emf), which is given by Faraday's law of electromagnetic induction as it applies to self-induction:
$$ 	ext{emf} = - L \frac{\Delta I}{\Delta t} $$
Where:
$ 	ext{emf} $ = induced voltage in volts (V)
$ L $ = self-inductance of the coil in henries (H)
$ \Delta I $ = change in current in amperes (A)
$ \Delta t $ = time interval in seconds (s) over which the current change occurs
Here, the problem gives us the following data:
$ 	ext{emf} = 0.1 \, 	ext{V} $
$ \Delta I = 2 \, 	ext{A} - (-2 \, 	ext{A}) = 4 \, 	ext{A} $
$ \Delta t = 0.2 \, 	ext{s} $
Substituting the given values into the formula, we get:
$$ 0.1 = -L \frac{4}{0.2} $$
Solving for $ L $, the equation becomes:
$$ 0.1 = -L \cdot 20 $$
Therefore:
$$ L = - \frac{0.1}{20} $$
Calculating the value of $ L $:
$$ L = -0.005 \, 	ext{H} $$
Or, expressing $ L $ in millihenries (mH):
$$ L = -5 \, 	ext{mH} $$
However, considering the absolute value (since inductance is a magnitude and cannot be negative in this context):
$$ L = 5 \, 	ext{mH} $$
Thus, the self-inductance of the coil is
5 mH
, which corresponds to
Option D
.

Answer

4 mH

Answer

2.5 mH

Answer

1 mH

Answer

5 mH

Electromagnetic Induction

In a coil, the current changes from $-2 \mathrm{~A} to +2 \mathrm{~A} in 0.2 \mathrm{~s} and induces an emf of 0.1 \mathrm{~V}$. The self inductance of the coil is :

Choose the correct answer: