Magnetic Effects & Magnetism

JEE Main level, single correct, medium

202 of 276

Question 202 of 276

A square loop of area 25cm225{cm}^{2} has a resistance of 10Ω10\Omega. The loop is placed in uniform magnetic field of magnitude 40.0T40.0T. The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in 1.01.0 sec, will be

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Answer: (B) 1.0×10−3J1.0\times {10}^{-3}J

As l2=25cm2{l}^{2}=25{cm}^{2}, then l=5cm=0.05ml=5cm=0.05m.
Given: t=1st=1s
Velocity of the square loop, v=0.051=0.05ms−1v=\dfrac{0.05}{1}=0.05m{s}^{-1}.
Induced current,
i=VR=BlvR=40×0.05×0.0510=0.01Ai=\dfrac{V}{R}=\dfrac{Blv}{R} =\dfrac{40\times 0.05\times 0.05}{10}=0.01A
Now force acting on the side of the square loop, F=Bil=40×0.01×0.05F=Bil=40\times 0.01\times 0.05
⇒F=0.02N\Rightarrow F=0.02N
Therefore, work done
W=Fl=0.02×l=0.02×0.05W=Fl =0.02\times l =0.02\times 0.05
⇒W=1×10−3J\Rightarrow W=1\times {10}^{-3}J

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