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QJEEM23Phfda0: JEE Main 2023 Physics - Class 12 JEEM23Phfda0 Solution: A massless square loop, of wire of resistance , supporting a…

A massless square loop, of wire of resistance 10Ω10 \Omega , supporting a mass of 1 g1 \mathrm{~g} , hangs vertically with one of its sides in a uniform magnetic field of 103G10^{3} \mathrm{G} , directed outwards in the shaded region. A dc voltage V\mathrm{V} is applied to the loop. For what value of V\mathrm{V} , the magnetic force will exactly balance the weight of the supporting mass of 1 g1 \mathrm{~g} ?

(If sides of the loop =10 cm, g=10 ms−2=10 \mathrm{~cm}, \mathrm{~g}=10 \mathrm{~ms}^{-2} )

JEE Main 2023 (Online) 30th January Morning Shift Physics - Magnetic Effect of Current Question 66 English

Multiple Choice Options

  • A 1 V
  • B 110\frac{1}{10} V
  • C 10 V
  • D 100 V

Step-by-Step Verified Solution

For balancing of force

∴Floop=weight\therefore F_{loop}=\mathrm{weight}

(VR)IB=mg\left( {{V \over R}} \right)IB = mg

(V10)×10100×(103×10−4)=(11000)×10\left( {{V \over {10}}} \right) \times {{10} \over {100}} \times ({10^3} \times {10^{ - 4}}) = \left( {{1 \over {1000}}} \right) \times 10

V=10V = 10 volts

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