Electromagnetic Induction

JEE Main level, single correct, medium

137 of 153

Question 137 of 153

When the current in a coil changes from 5 A\text{5 A} to 2 A\text{2 A} in 0.1 s\text{0}\text{.1 s}, an average voltage of 50V\text{50} \text{V} is produced. The self-inductance of the coil is

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Answer: (A) 1.67 H\text{1}\text{.67 H}

Magnetic flux

ϕ=LI⇒∆ϕ∆t=L∆I∆t\phi =LI\Rightarrow \dfrac{∆\phi }{∆t}=L\dfrac{∆I}{∆t}

∴(ϵind)average=∣∆ϕ∆t∣=L∣∆I∆t∣∴{\left({\epsilon }_{ind}\right)}_{average}=\left|\dfrac{∆\phi }{∆t}\right|=L\left|\dfrac{∆I}{∆t}\right|

⇒50=L×5−20.1\Rightarrow 50=L\times \dfrac{5-2}{0.1}

⇒53=L\Rightarrow \dfrac{5}{3}=L

⇒L=1.67H\Rightarrow L=1.67H

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