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JEE Main 202330 JanuaryMorning Shift
Exam
JEE Main
Year
2023
Date
30 January
Shift
Morning Shift
Subject
Physics
Chapter
Magnetic Effects of Current

JEE Main 2023 · 30 January · Morning Shift · Physics · Magnetic Effects of Current

A massless square loop, of wire of resistance 10Ω , supporting a...

A massless square loop, of wire of resistance 10Ω10Ω, 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})

A massless square loop, of wire of resistance 10Ω , supporting a mass of 1g , hangs vertically with one of its sides in.

Options

  1. A110V\dfrac{1}{10}V
  2. B100V100V
  3. C1V1V
  4. D10V10V

Correct answer

D. 10V10V

Step-by-step solution

For balancing of force,

Magnetic force (Fm)=\left({F}_{m}\right) =weight of supporting mass \left(mg\right).</p><p>.</p><p>∴ ILB = mg</p><p></p><p>∴ \left(\dfrac{V}{R}\right)LB = mg\left(∵ \text{From Ohm's law}, I = \dfrac{V}{R}\right)</p><p></p><p>∴ V = \dfrac{mgR}{LB}</p><p></p><p>= \dfrac{\left(1 \times {10}^{- 3}kg\right)\left(10m{s}^{- 2}\right)\left(10Ω\right)}{\left(0 . 1m\right)\left({10}^{3} \times {10}^{- 4}T\right)} = 10V$

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