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(D) The magnetic force $F_m$ acting on the side of the loop inside the magnetic field must balance the weight $mg$ of the mass.$F_m = mg$Since $F_m =

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$10\,V$

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(D) The magnetic force $F_m$ acting on the side of the loop inside the magnetic field must balance the weight $mg$ of the mass.$F_m = mg$Since $F_m = ILB$,we have $ILB = mg$.Using Ohm's law,$I = \frac{V}{R}$,so the equation becomes $\left(\frac{V}{R}ight)LB = mg$.Rearranging for $V$,we get $V = \frac{mgR}{LB}$.Given values:$m = 1\,g = 10^{-3}\,kg$$g = 10\,m/s^2$$R = 10\,\Omega$$L = 10\,cm = 0.1\,m$$B = 10^3\,G = 10^3 	imes 10^{-4}\,T = 0.1\,T$Substituting these values:$V = \frac{(10^{-3}\,kg)(10\,m/s^2)(10\,\Omega)}{(0.1\,m)(0.1\,T)} = \frac{10^{-1}}{10^{-2}} = 10\,V$.

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