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← JEE Main 2014 · 19 Apr · Shift 1
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JEE Main 2014 · 19 Apr · Shift 1, Physics question 19

A square frame of side 10 cm and a long straight wire carrying current 1 A are in the plane of the paper. Starting from close to the wire, the frame moves towards the right with a constant speed of 10 ms−1^{-1} (see figure). The e.m.f induced at the time the left arm of the frame is at x=10x=10 cm from the wire is :

A long vertical wire carrying I = 1 A upward; a square frame of side 10 cm, its left arm a distance x from the wire, moves to the right with velocity v.
Choose one option
Answer and worked solution

Official answer: B

The near and far arms (10 cm long) cut field lines at xx and x+ax+a:
ε=μ0Iav2π(1x−1x+a)=2×10−7×1×0.1×10×(10−5)=1 μV.\varepsilon=\frac{\mu_0Iav}{2\pi}\left(\frac1x-\frac1{x+a}\right)=2\times10^{-7}\times1\times0.1\times10\times(10-5)=1\ \mu\text{V}.

Topics:Motional EMFBiot-Savart law
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