Magnetic Effects & Magnetism

JEE Main level, single correct, medium

201 of 276

Question 201 of 276

A massless square loop, of wire of resistance 10Ω10\Omega, supporting a mass of 1g1g, hangs vertically with one of its sides in a uniform magnetic field of 103G{10}^{3}G, directed outwards in the shaded region. A dc voltage VV is applied to the loop. For what value of VV, the magnetic force will exactly balance the weight of the supporting mass of 1g1g?
(If sides of the loop =10cm,g=10ms−2=10cm,g=10m{s}^{-2})
.

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Answer: (D) 10V10V

For balancing of force,
Magnetic force (Fm)=\left({F}_{m}\right)=weight of supporting mass (mg)\left(mg\right).
∴ILB=mg∴ILB=mg
∴(VR)LB=mg(∵From Ohm’s law,I=VR)∴\left(\dfrac{V}{R}\right)LB=mg\left(∵\text{From Ohm's law},I=\dfrac{V}{R}\right)
∴V=mgRLB∴V=\dfrac{mgR}{LB}
=(1×10−3kg)(10ms−2)(10Ω)(0.1m)(103×10−4T)=10V=\dfrac{\left(1\times {10}^{-3}kg\right)\left(10m{s}^{-2}\right)\left(10\Omega \right)}{\left(0.1m\right)\left({10}^{3}\times {10}^{-4}T\right)}=10V

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