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 is ___

To determine the magnetic moment of a rotating charged disc, we need to consider how the charge distribution interacts with its motion. When a charge is uniformly distributed over a rotating disc, it creates a magnetic moment due to the motion of the charges. Let's break this down step by step.

Understanding Magnetic Moment

The magnetic moment (\( \mu \)) is a vector quantity that represents the strength and direction of a magnetic source. For a current loop, it can be calculated using the formula:

• \( \mu = I \cdot A \)

where \( I \) is the current flowing through the loop and \( A \) is the area of the loop. In our case, the rotating disc acts like a series of current loops due to the motion of the charge.

Calculating the Current

First, we need to find the effective current generated by the rotating charge. The total charge \( Q \) is uniformly distributed over the surface of the disc. The surface area \( A \) of the disc is given by:

• \( A = \pi R^2 \)

As the disc rotates with an angular velocity \( \omega \), each infinitesimal charge element \( dq \) on the disc moves in a circular path. The linear velocity \( v \) of a point at a distance \( r \) from the center is:

• \( v = r \cdot \omega \)

Finding the Effective Current

The current \( I \) can be thought of as the charge passing through a point per unit time. For a small ring of radius \( r \) and thickness \( dr \), the charge \( dq \) on that ring is:

• \( dq = \frac{Q}{\pi R^2} \cdot 2\pi r \cdot dr = \frac{2Qr}{R^2} \cdot dr \)

The current \( dI \) due to this charge moving around the ring is given by:

• \( dI = \frac{dq}{T} \)

where \( T \) is the period of rotation. The period \( T \) can be expressed as:

• \( T = \frac{2\pi}{\omega} \)

Substituting this into the current equation gives:

• \( dI = \frac{2Qr}{R^2} \cdot dr \cdot \frac{\omega}{2\pi} = \frac{Qr\omega}{\pi R^2} \cdot dr \)

Calculating the Magnetic Moment

Now, we can find the magnetic moment for the entire disc by integrating \( dI \) over the radius of the disc:

• \( \mu = \int_0^R dI \cdot A = \int_0^R \left(\frac{Qr\omega}{\pi R^2}\right) \cdot \pi r^2 \, dr \)

This simplifies to:

• \( \mu = \frac{Q\omega}{R^2} \int_0^R r^3 \, dr = \frac{Q\omega}{R^2} \cdot \frac{R^4}{4} = \frac{QR^2\omega}{4} \)

Final Expression for Magnetic Moment

Thus, the magnetic moment of the rotating non-conducting disc is:

• \( \mu = \frac{QR^2\omega}{4} \)

This result shows how the magnetic moment depends on the total charge, the radius of the disc, and the angular velocity. The larger the charge or the faster the rotation, the greater the magnetic moment produced by the disc.