Electromagnetic Induction

JEE Main level, numerical answer, medium

55 of 153

Question 55 of 153

A 10Ω,20mH10\Omega ,20mH coil carrying constant current is connected to a battery of 20V20V through a switch. Now after switch is opened current becomes zero in 100μs100\mu s. The average e.m.f. induced in the coil is _____ VV.

Show answer and solution

Answer: 400

The initial current in the coil will be, i=VR=2010=2Ai=\dfrac{V}{R}=\dfrac{20}{10}=2A.
Average e.m.f. developed in the coil will be,
ϵ=−∆ϕ∆t=0−Lit=20×10−3×2100×10−6=400V\epsilon =-\dfrac{∆\phi }{∆t}=\dfrac{0-Li}{t}=\dfrac{20\times {10}^{-3}\times 2}{100\times {10}^{-6}}=400V.

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Electromagnetic induction simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Electromagnetic Induction (Induced Electric Field), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.