Question 309: Electromagnetic Induction

Question

A current $$I=10 \sin (100 \pi t)$$ ampere, is passed in a coil which induces a maximum emf $$5 \pi$$ volt in neighbouring coil. The mutual inductance of two coils is

Accepted Answer

To determine the mutual inductance between the two coils, we can use Faraday's Law of Induction. According to Faraday's Law, the electromotive force (emf) induced in a coil is given by:
$$ \mathcal{E} = -M \frac{dI}{dt} $$
where:
$$\mathcal{E}$$ is the induced emf,
$$M$$ is the mutual inductance,
$$I$$ is the current in the primary coil.
Given the current in the primary coil:
$$ I = 10 \sin (100 \pi t) \, 	ext{A} $$
First, we need to compute the derivative of the current with respect to time:
$$ \frac{dI}{dt} = \frac{d}{dt} (10 \sin (100 \pi t)) = 10 \cdot 100 \pi \cos (100 \pi t) = 1000 \pi \cos (100 \pi t) $$
We are given that the maximum emf induced in the neighbouring coil is:
$$ \mathcal{E}_{	ext{max}} = 5 \pi \, 	ext{V} $$
The maximum value of cosine function is 1. Therefore, the maximum emf is induced when:
$$ \mathcal{E}_{	ext{max}} = M \cdot (1000 \pi) $$
Setting the given maximum emf equal to this expression, we get:
$$ 5 \pi = M \cdot 1000 \pi $$
Solving for $$M$$:
$$ M = \frac{5 \pi}{1000 \pi} = \frac{5}{1000} = 0.005 \, 	ext{H} $$
Converting Henry to milliHenry (1 H = 1000 mH):
$$ M = 0.005 \, 	ext{H} = 5 \, 	ext{mH} $$
Therefore, the mutual inductance of the two coils is:
Option A: $$5 \mathrm{~mH}$$

Answer

$$5 \mathrm{~mH}$$

Answer

$$10 \mathrm{~mH}$$

Answer

$$15 \mathrm{~mH}$$

Answer

$$25 \mathrm{~mH}$$

Electromagnetic Induction

A current $I=10 \sin (100 \pi t) ampere, is passed in a coil which induces a maximum emf 5 \pi$ volt in neighbouring coil. The mutual inductance of two coils is

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