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Magnetics question

2018 · 16 Apr · Shift 1 · Q53
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Magnetics question

2018 · 16 Apr · Shift 1 · Q53

JEE MainPhysicsMagneticsMCQ+4 / −1
A charge q is spread uniformly over an insulated loop of radius r. If it is rotated with an angular velocity ω\omegaω with resect to normal axis then the magnetic moment of the loop is :
  1. A
    q ω\omegaω r2
  2. B
    43{4 \over 3}34​ q ω\omegaω r2
  3. C
    32{3 \over 2}23​ q ω\omegaω r2
  4. D
    12{1 \over 2}21​ q ω\omegaω r2
View written solutionFree

Correct answer: D

  1. Magnetic moment of a current loop

If a charge distribution rotates in a circle, it constitutes a current. The magnetic moment is

μ=IA\mu = I Aμ=IA

where:

  • III = current
  • A=πr2A = \pi r^2A=πr2 = area of the loop
  1. Find the current

The total charge on the insulated loop is qqq, uniformly distributed. If the loop rotates with angular velocity ω\omegaω, then one full revolution takes time

T=2πωT = \frac{2\pi}{\omega}T=ω2π​

So the equivalent current is

I=qT=qω2πI = \frac{q}{T} = \frac{q\omega}{2\pi}I=Tq​=2πqω​
  1. Compute magnetic moment

Now,

μ=IA=qω2π⋅πr2\mu = IA = \frac{q\omega}{2\pi} \cdot \pi r^2μ=IA=2πqω​⋅πr2 μ=12qωr2\mu = \frac{1}{2} q\omega r^2μ=21​qωr2
  1. Match with the options

Thus the magnetic moment is

12qωr2\boxed{\frac{1}{2} q\omega r^2}21​qωr2​

This matches Option D.

  1. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.

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