A closed current-carrying loop having a current I is having area A. Magnetic moment of this loop is defined as `vec(mu) = vec(IA)` where direction of area vector is towards the observer if current is flowing in anticlockwise direction with respect to the observer. If this loop is placed in a uniform magnetic field `vec(B)`, then torque acting on the loop is given by `vec(tau) = vec(mu) xx vec(B)`. Now answer the following questions:
Let ring in the above question is having a radius R and a charge Q is uniformly distributed over it. Ring is rotated with a constant angular velocity `(omega)` as mentioned above.
Torque acting on the ring due to magnetic force is

Options

A. `(QR^(2)omega B)/(2)`
B. `(QR^(2)omega B)/(2 omega)`
C. `(omega R^(2)B)/(2 pi)`
D. None of the above

Explanation

To solve the problem, we need to calculate the torque acting on a rotating ring with a uniformly distributed charge in a uniform magnetic field. Here's a step-by-step solution:

### Step 1: Understand the Magnetic Moment
The magnetic moment (μ) of a current-carrying loop is given by the formula:
\[ \vec{\mu} = \vec{IA} \]
where \( I \) is the current and \( A \) is the area of the loop. For a ring of radius \( R \), the area \( A \) is:
\[ A = \pi R^2 \]

### Step 2: Relate Charge to Current
The current \( I \) can be expressed in terms of the charge \( Q \) and the time period \( T \) of rotation:
\[ I = \frac{Q}{T} \]
The time period \( T \) for one complete rotation at angular velocity \( \omega \) is:
\[ T = \frac{2\pi}{\omega} \]
Thus, we can substitute \( T \) into the expression for current:
\[ I = \frac{Q \omega}{2\pi} \]

### Step 3: Calculate the Magnetic Moment
Now substituting \( I \) and \( A \) into the magnetic moment formula:
\[ \mu = I \cdot A = \left(\frac{Q \omega}{2\pi}\right) \cdot (\pi R^2) = \frac{Q \omega R^2}{2} \]

### Step 4: Calculate the Torque
The torque \( \tau \) acting on the loop in a magnetic field \( \vec{B} \) is given by:
\[ \vec{\tau} = \vec{\mu} \times \vec{B} \]
The magnitude of the torque can be expressed as:
\[ \tau = \mu B \sin \theta \]
where \( \theta \) is the angle between the magnetic moment vector and the magnetic field vector.

### Step 5: Substitute the Magnetic Moment
Substituting the expression for \( \mu \) into the torque equation:
\[ \tau = \left(\frac{Q \omega R^2}{2}\right) B \sin \theta \]

### Step 6: Special Cases
- If the magnetic field \( \vec{B} \) is perpendicular to the area vector (i.e., \( \theta = 90^\circ \)), then \( \sin \theta = 1 \):
\[ \tau = \frac{Q \omega R^2 B}{2} \]
- If the magnetic field is at some other angle, the torque will be less than this maximum value.

### Final Expression
Thus, the torque acting on the ring due to the magnetic force is:
\[ \tau = \frac{Q \omega R^2 B}{2} \sin \theta \]