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A charge q is spread uniformly over an insulated loop of radius r. If it is rotated
with an angular velocity ω with respect to normal axis then the magnetic moment of
the loop is :

  • Option 1)

    q\omega r^{2}

  • Option 2)

    4/3 q\omega r^{2}

  • Option 3)

    3/2 q\omega r^{2}

  • Option 4)

    1/2 q\omega r^{2}

 

Answers (1)

best_answer

As we have learned

Magnetic moment (M) -

M=NiA

- wherein

N-number of turns in the coil 

i-current throughout the coil 

A-area of the coil 

 

for any charge distribution

 \mu /L=\left ( \frac{q}{2m} \right )            or             \mu = \frac{1}{2}\left ( \frac{q}{2m} \right )L

L= IW= mr^{2}w

\mu = 1/2 qr^{2}w

 

 

 

 


Option 1)

q\omega r^{2}

This is incorrect

Option 2)

4/3 q\omega r^{2}

This is incorrect

Option 3)

3/2 q\omega r^{2}

This is incorrect

Option 4)

1/2 q\omega r^{2}

This is incorrect

Posted by

Avinash

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