JEE Main 2014 — Physics Electromagnetism

2014mcqhard

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

JEE Main 2014 Physics, Electromagnetic Induction — question figure

Official previous-year question

Held on 19 Apr 2014 · Verified 6 Jul 2026.

Options

A
$0.75\mu V$
B
$1 \mu V$
C
$2 \mu V$
D
$0.5\mu V$

Did you get this right?

Solution

JEE Main 2014 Physics, Electromagnetic Induction — solution figure 1

$d\phi =\frac{{\mu }_{0}i}{2\pi y}ady$

$\phi =\frac{{\mu }_{0}ia}{2\pi }\int _{x}^{x+a}\frac{dy}{y}=\frac{{\mu }_{0}ia}{2\pi }[ln(x+a)-lnx]$

$\text{E.m.f}=-\frac{d\phi }{dt}=-\frac{{\mu }_{0}ia}{2\pi }[\frac{1}{x+a}\frac{dx}{dt}-\frac{1}{x}\frac{dx}{dt}]$

$=\frac{{\mu }_{0}ia}{2\pi }\frac{a.v}{x(x+a)}=\frac{{\mu }_{0}}{2\pi }.\frac{i{a}^{2}v}{x(x+a)}=2\times {10}^{-7}\times \frac{1\times ({0.1}^{2}\times 10)}{0.1\times 0.2}=1\mu V$

Alternative method:

Consider two sides which are perpendicular to the velocity as case of motional e.m.f.

JEE Main 2014 Physics, Electromagnetic Induction — solution figure 2

$E={E}_{1}-{E}_{2}=({B}_{1}–{B}_{2})Vl$

$=\frac{{\mu }_{0}I}{2\pi }(\frac{1}{x}-\frac{1}{x+a})Vl$

$=2\times {10}^{-7}\times 1(\frac{1}{0.1}-\frac{1}{0.2})\times 10\times 0.1$

$={10}^{-6}V=1 \mu V$

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Solution

JEE Main 2014 Physics, Electromagnetic Induction — solution figure 1

$d\phi =\frac{{\mu }_{0}i}{2\pi y}ady$

$\phi =\frac{{\mu }_{0}ia}{2\pi }\int _{x}^{x+a}\frac{dy}{y}=\frac{{\mu }_{0}ia}{2\pi }[ln(x+a)-lnx]$

$\text{E.m.f}=-\frac{d\phi }{dt}=-\frac{{\mu }_{0}ia}{2\pi }[\frac{1}{x+a}\frac{dx}{dt}-\frac{1}{x}\frac{dx}{dt}]$

$=\frac{{\mu }_{0}ia}{2\pi }\frac{a.v}{x(x+a)}=\frac{{\mu }_{0}}{2\pi }.\frac{i{a}^{2}v}{x(x+a)}=2\times {10}^{-7}\times \frac{1\times ({0.1}^{2}\times 10)}{0.1\times 0.2}=1\mu V$

Alternative method:

Consider two sides which are perpendicular to the velocity as case of motional e.m.f.

JEE Main 2014 Physics, Electromagnetic Induction — solution figure 2

$E={E}_{1}-{E}_{2}=({B}_{1}–{B}_{2})Vl$

$=\frac{{\mu }_{0}I}{2\pi }(\frac{1}{x}-\frac{1}{x+a})Vl$

$=2\times {10}^{-7}\times 1(\frac{1}{0.1}-\frac{1}{0.2})\times 10\times 0.1$

$={10}^{-6}V=1 \mu V$

JEE Main 2014 — Physics Electromagnetism

Options

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Solution

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JEE Main 2014 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism