Medium MCQ +4 / -1 PYQ · JEE Mains 2023

A square loop of area 25 cm$^2$ has a resistance of 10 $\Omega$. The loop is placed in uniform magnetic field of magnitude 40.0 T. 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 1.0 sec, will be

  1. A $\mathrm{1.0\times10^{-3}~J}$ Correct answer
  2. B $\mathrm{5\times10^{-3}~J}$
  3. C $\mathrm{2.5\times10^{-3}~J}$
  4. D $\mathrm{1.0\times10^{-4}~J}$

Solution

<p>From energy conservation</p> <p>Work done to pull the loop out = Energy is lost in the resistance</p> <p>Emf in the loop $$ = {{d\phi } \over {dt}} = {{B \times A} \over t} = {{40 \times 25 \times {{10}^{ - 4}}} \over {1s}} = 0.1\,V$$</p> <p>Energy lost $= {{em{f^2}} \over R} = {{{{(0.1)}^2}} \over {10}} = {10^{ - 3}}\,J$</p>

About this question

Subject: Physics · Chapter: Electromagnetic Induction · Topic: Motional EMF

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