Electromagnetic Induction

JEE Main level, single correct, medium

12 of 153

Question 12 of 153

A 12V12V battery connected to a coil of resistance 6Ω6Ω through a switch, drives a constant current in the circuit. The switch is opened in 1ms1ms. The emf induced across the coil is 20V20V. The inductance of the coil is :

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Answer: (A) 10mH10mH

The current (i)\left(i\right) through the circuit is given by
i=VR=12V6Ω=2A\begin{matrix}i & = & \dfrac{V}{R} \\ & = & \dfrac{12V}{6\Omega } \\ & = & 2A\end{matrix}
The formula to calculate the induced emf in the coil can be written as
ϵ=Ldidt...(1)\epsilon =L\dfrac{di}{dt}...\left(1\right)
Substitute the values of the known parameters into equation (1) and solve to calculate the inductance of the coil.
20V=L×2A1×10−3s⇒L=20V×1×10−3s2A=10mH\begin{matrix}20V & = & L\times \dfrac{2A}{1\times {10}^{-3}s} \\ & \Rightarrow & L=\dfrac{20V\times 1\times {10}^{-3}s}{2A} \\ & = & 10mH\end{matrix}

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