A massless square loop of wire of resistance ' $R$ ' supporting a mass ' M ' hangs vertically with one of…

A massless square loop of wire of resistance ' $R$ ' supporting a mass ' M ' hangs vertically with one of its sides in a uniform magnetic field ' $B$ ' directed outwards in the shaded region. A d.c. voltage ' V ' is applied to the loop. For what value of ' $V$ ' the magnetic force will exactly balance the weight of the supporting mass ' $M$ '? (side of loop $=\mathrm{L}, \mathrm{g}=$ acceleration due to gravity)
  1. $\frac{\mathrm{Mg}}{\mathrm{LBR}}$
  2. $\frac{\mathrm{LB}}{\mathrm{MgR}}$
  3. $\frac{\mathrm{MgR}}{\mathrm{LB}}$
  4. $\frac{\mathrm{LR}}{\mathrm{MgB}}$

Solution

$\begin{array}{ll} & \mathrm{F}=\mathrm{Mg} \\ \therefore \quad & \mathrm{BIL}=\mathrm{Mg} \\ & \text { According to Ohm's law, } \mathrm{V}=\mathrm{IR} \\ \therefore \quad & \mathrm{B}\left(\frac{\mathrm{V}}{\mathrm{R}}\right) \mathrm{L}=\mathrm{Mg} \\ \therefore \quad & \mathrm{V}=\frac{\mathrm{MgR}}{\mathrm{BL}}\end{array}$

Asked in: MHT CET 2024 (04 May Shift 1)

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