Question

# A non conducting disc of radius $$R$$, charge $$q$$ is rotating about an axis passing through its centre and perpendicular to its plane with an angular velocity $$\omega$$, charge $$q$$ is uniformly distributed over its surface. The magnetic moment of the disc is :

A
14qωR2
B
12qωR
C
qωR
D
12qωR2

Solution

## The correct option is B $$\frac{1}{4}q\omega R^{2}$$Charge per unit area$$(\sigma )=\dfrac{q}{\pi R^2}$$Charge on ring of width $$dr$$$$=\sigma \cdot2\pi rdr$$Magnetic moment of this ring of width dr$$d\mu=i\cdot A$$$$\displaystyle \int d\mu=\int_{0}^{R}T\cdot2\pi rdr\dfrac{w}{2\pi}\cdot\pi r^2$$$$=\mu=\dfrac{qwR^2}{4}$$PhysicsNCERTStandard XII

Suggest Corrections

0

Similar questions
View More

Same exercise questions
View More

People also searched for
View More