JEE Main On 16 April 2018 Question 1
Question: A charge q is spread uniformly over an insulated loop of radius. If it is rotated with an angular velocity $ \omega $ with respect to normal axis then the magnetic moment of the loop is [JEE Main 16-4-2018]
Options:
A) $ \frac{1}{2}q\omega r^{2} $
B) $ \frac{4}{3}q\omega r^{2} $
C) $ \frac{3}{2}q\omega r^{2} $
D) $ q\omega r^{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
Let us take an element at an angle $ \theta $ subtending an angle $ d\theta $
The charge the element has can be written as $ dq=\frac{q}{2\pi }d\theta $
We know that $ i=\frac{dq}{dt}=\frac{q}{2\pi }\times \frac{d\theta }{dt} $
The time dt can be written as $ dt=\frac{d\theta }{w} $ Hence $ i=\frac{qw}{2\pi } $
magnetic moment = iA Hence Magnetic moment $ M=\frac{qw}{2\pi }\times \pi r^{2}=\frac{qwr^{2}}{2} $
