Question 264: Electromagnetic Induction

Question

Two coils are kept near each other. When no current passess through first coil and current in the $2^{	ext {nd }}$ coil increases at the rate $10 \mathrm{~A} / \mathrm{s}$, the e.m.f. in the $1^{\mathbb{P}}$ coil is 20 mV . When no current passes through $2^{	ext {nd }}$ coil and 3.6 A current passes through $1^2$ coil the flux linkage in coil 2 is

Accepted Answer

When analyzing the scenario of two coils placed near each other, the mutual inductance ($M$) plays a key role in determining the electromotive force (e.m.f.) induced in one coil due to a change in current of the other. The e.m.f. ($\varepsilon$) induced due to the rate of change of current in the second coil is given by:
$$ \varepsilon = -M \frac{dI_2}{dt} $$
Given that $\varepsilon = 20 \, 	ext{mV} = 20 	imes 10^{-3} \, 	ext{V}$ and the rate of change of current in the second coil $\frac{dI_2}{dt} = 10 \, 	ext{A/s}$, we can rearrange the formula to solve for the mutual inductance $M$:
$$ M = -\frac{\varepsilon}{\frac{dI_2}{dt}} = -\frac{20 	imes 10^{-3}}{10} = 2 	imes 10^{-3} \, 	ext{H} $$
Now, consider the situation where no current passes through the second coil, and a current of 3.6 A passes through the first coil. The flux linkage ($\Phi_{21}$) in the second coil due to the current in the first coil can be found using:
$$ \Phi_{21} = M I_1 $$
Substitute $M = 2 	imes 10^{-3} \, 	ext{H}$ and $I_1 = 3.6 \, 	ext{A}$:
$$ \Phi_{21} = (2 	imes 10^{-3} \, 	ext{H})(3.6 \, 	ext{A}) = 7.2 	imes 10^{-3} \, 	ext{Wb} $$
Therefore, the correct answer is:
Option D: $7.2 	imes 10^{-3} \, 	ext{Wb}$

Answer

$1.2 	imes 10^{-3} \mathrm{~Wb}$

Answer

$1.8 	imes 10^{-3} \mathrm{~Wb}$

Answer

$3.6 	imes 10^{-3} \mathrm{~Wb}$

Answer

$7.2 	imes 10^{-3} \mathrm{~Wb}$

Electromagnetic Induction

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