JEE MAIN - Physics (2023 - 29th January Evening Shift - No. 4)
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
$$\mathrm{1.0\times10^{-3}~J}$$
$$\mathrm{5\times10^{-3}~J}$$
$$\mathrm{2.5\times10^{-3}~J}$$
$$\mathrm{1.0\times10^{-4}~J}$$
Explanation
From energy conservation
Work done to pull the loop out = Energy is lost in the resistance
Emf in the loop $$ = {{d\phi } \over {dt}} = {{B \times A} \over t} = {{40 \times 25 \times {{10}^{ - 4}}} \over {1s}} = 0.1\,V$$
Energy lost $$ = {{em{f^2}} \over R} = {{{{(0.1)}^2}} \over {10}} = {10^{ - 3}}\,J$$
Comments (0)
