MHT CET · Physics · Magnetic Effects of Current
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)
- A \(\frac{\mathrm{Mg}}{\mathrm{LBR}}\)
- B \(\frac{\mathrm{LB}}{\mathrm{MgR}}\)
- C \(\frac{\mathrm{MgR}}{\mathrm{LB}}\)
- D \(\frac{\mathrm{LR}}{\mathrm{MgB}}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{MgR}}{\mathrm{LB}}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll} & \mathrm{F}=\mathrm{Mg} \\ \therefore \quad & \mathrm{BIL}=\mathrm{Mg} \\ & \text { According to Ohm's law, } \mathrm{V}=\mathrm{IR} \\ \therefore \quad & \mathrm{B}\left(\frac{\mathrm{V}}{\mathrm{R}}\right) \mathrm{L}=\mathrm{Mg} \\ \therefore \quad & \mathrm{V}=\frac{\mathrm{MgR}}{\mathrm{BL}}\end{array}\)
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