A massless square loop, of wire of resistance 10Ω, supporting a mass of 1 g, hangs vertically with one of its sides in a uniform magnetic field of 10³G , directed outwards in the shaded region. A dc voltage V is applied to the loop. For what value of V, the magnetic force will exactly balance the weight of the supporting mass of 1g?
(If sides of the loop=10 cm, g=10ms^-2)

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100 V

$\frac{1}{10}$ V

1 V

10V

answer is D.

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Detailed Solution

$\begin{array}{l} 1 Gauss = 10^{- 4} T \\ \Rightarrow B = 10^{- 1} T \\ BIL = mg [in equilibrium] \\ 10^{- 1} (\frac{V}{10}) (0.1) = 10^{- 3} \times 10 \\ V = 10 V \end{array}$

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