JEE MAIN - Physics (2018 - 16th April Morning Slot - No. 6)
A charge q is spread uniformly over an insulated loop of radius r. If it is rotated with an angular velocity $$\omega $$ with resect to normal axis then the magnetic moment of the loop is :
q $$\omega $$r2
$${4 \over 3}$$ q $$\omega $$r2
$${3 \over 2}$$ q $$\omega $$r2
$${1 \over 2}$$ q $$\omega $$r2
Forklaring
Magnetic moment,
$$\mu $$ = I A
= $${q \over T}\left( {\pi {r^2}} \right)$$
= $${q \over {2\pi /\omega }}\left( {\pi {r^2}} \right)$$
= $${{qw} \over {2\pi }}$$ $$\left( {\pi {r^2}} \right)$$
= $${1 \over 2}$$ q$$\omega $$r2
$$\mu $$ = I A
= $${q \over T}\left( {\pi {r^2}} \right)$$
= $${q \over {2\pi /\omega }}\left( {\pi {r^2}} \right)$$
= $${{qw} \over {2\pi }}$$ $$\left( {\pi {r^2}} \right)$$
= $${1 \over 2}$$ q$$\omega $$r2