JEE MAIN - Physics (2023 - 30th January Morning Shift - No. 1)
A massless square loop, of wire of resistance $$10 \Omega$$, supporting a mass of $$1 \mathrm{~g}$$, hangs vertically with one of its sides in a uniform magnetic field of $$10^{3} \mathrm{G}$$, directed outwards in the shaded region. A dc voltage $$\mathrm{V}$$ is applied to the loop. For what value of $$\mathrm{V}$$, the magnetic force will exactly balance the weight of the supporting mass of $$1 \mathrm{~g}$$ ?
(If sides of the loop $$=10 \mathrm{~cm}, \mathrm{~g}=10 \mathrm{~ms}^{-2}$$)
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1 V
$$\frac{1}{10}$$V
10 V
100 V
Explanation
For balancing of force
$$\therefore F_{loop}=\mathrm{weight}$$
$$\left( {{V \over R}} \right)IB = mg$$
$$\left( {{V \over {10}}} \right) \times {{10} \over {100}} \times ({10^3} \times {10^{ - 4}}) = \left( {{1 \over {1000}}} \right) \times 10$$
$$V = 10$$ volts
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