0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

# A charge q is distributed uniformly around a thin ring of radius r. The ring is rotating about an axis through its center and perpendicular to its plane, at an angular speed $$\omega$$. (a) Show that the magnetic moment due to the rotating charge has a magnitude of $$\mu=\frac{1}{2}q\omega r^2$$ . (b) What is the direction of this magnetic moment if the charge is positive?

Solution
Verified by Toppr

#### (a) The period of rotation is $$T = 2π/ω$$, and in this time all the charge passes any fixedpoint near the ring. The average current is $$i = q/T = qω/2π$$ and the magnitude of the magnetic dipole moment is $$\mu=iA=\dfrac{q\omega}{2\pi}\pi r^2=\dfrac{1}{2}q\omega r^2$$ (b) We curl the fingers of our right hand in the direction of rotation. Since the charge is positive, the thumb points in the direction of the dipole moment. It is the same as the direction of the angular momentum vector of the ring.

2
Similar Questions
Q1
A charge q is distributed uniformly around a thin ring of radius r. The ring is rotating about an axis through its center and perpendicular to its plane, at an angular speed $$\omega$$. (a) Show that the magnetic moment due to the rotating charge has a magnitude of $$\mu=\frac{1}{2}q\omega r^2$$ . (b) What is the direction of this magnetic moment if the charge is positive?
View Solution
Q2

A non conducting disc of radius R is rotating about an axis passing through its centre and perpendicular to its plane with an angular velocity ω. Charge q is uniformly distributed over its surface. The magnetic moment of the disc is

View Solution
Q3
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 ω, charge q is uniformly distributed over its surface. The magnetic moment of the disc is :
View Solution
Q4
A ring of radius R, made of an insulating material carries a charge Q uniformly distributed on it. If the ring rotates about the axis passing through its centre and normal to plane of the ring with constant angular speed ω, then the magnitude of the magnetic moment of the ring will be?
View Solution
Q5
Consider a non-conducting ring of radius r and mass m that has a total charge q distributed uniformly on it. The ring is rotated about its axis with an angular speed ω. (a) Find the equivalent electric current in the ring. (b) Find the magnetic moment µ of the ring. (c) Show that $\mathrm{\mu }=\frac{q}{2m}$ l, where l is the angular momentum of the ring about its axis of rotation.
View Solution