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  4. A massless square loop, of wire of resistance 10 Ω, supporting a mass of 1 g, hangs vertically with one of its sides in a uniform magnetic field of 103 G, directed outwards in the shaded region. A de 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 1 g ? (If sides of the loop =10 cm , g =10 ms -2 )

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Q. A massless square loop, of wire of resistance $10\, \Omega$, supporting a mass of $1 \, g$, hangs vertically with one of its sides in a uniform magnetic field of $10^3 G$, directed outwards in the shaded region. A de 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 $1 g$ ? (If sides of the loop $=10 \, cm , g =10\, ms ^{-2}$ )Physics Question Image

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Solution:

$ F _{ m }= mg$
$\therefore \text { ILB }= mg $
$ \therefore \left(\frac{ V }{ R }\right) LB = mg$
$ \therefore V =\frac{ mgR }{ LB } $
$ =\frac{\left(1 \times 10^{-3} kg \right)\left(10 m / s ^2\right)(10 \Omega)}{(0.1 m )\left(10^3 \times 10^{-4} T \right)}=10 \,V $