A charge q is spread uniformly over an insulated loop of radius r. If it is rotated with an angular velocity `omega` with respect to normal axis then the magnetic moment of the loop is

Options

A. `1/2qomegar^2`
B. `4/3qomegar^2`
C. `3/2qomegar^2`
D. `qomegar^2`

Explanation

To find the magnetic moment of a uniformly charged loop rotating with an angular velocity \(\omega\), we can follow these steps:

### Step 1: Understand the charge distribution
The charge \(q\) is uniformly distributed over a loop of radius \(r\).

**Hint:** Recall that uniform distribution means the charge density is constant throughout the loop.

### Step 2: Determine the current \(I\)
When the loop rotates, it generates a current. The current \(I\) can be defined as the charge passing a point in the loop per unit time. The time period \(T\) for one complete rotation can be expressed as:

\[ T = \frac{2\pi}{\omega} \]

Thus, the current \(I\) can be calculated as:

\[ I = \frac{q}{T} = \frac{q}{\frac{2\pi}{\omega}} = \frac{q \omega}{2\pi} \]

**Hint:** Remember that current is the charge per time.

### Step 3: Calculate the area \(A\) of the loop
The area \(A\) of a circular loop is given by:

\[ A = \pi r^2 \]

**Hint:** Use the formula for the area of a circle, which is \(\pi r^2\).

### Step 4: Calculate the magnetic moment \(\mu\)
The magnetic moment \(\mu\) of a current loop is given by the formula:

\[ \mu = I \cdot A \]

Substituting the expressions for \(I\) and \(A\):

\[ \mu = \left(\frac{q \omega}{2\pi}\right) \cdot \left(\pi r^2\right) \]

### Step 5: Simplify the expression
Now, simplifying the expression for \(\mu\):

\[ \mu = \frac{q \omega}{2\pi} \cdot \pi r^2 = \frac{q \omega r^2}{2} \]

**Hint:** Look for common factors that can be canceled out in the expression.

### Final Result
Thus, the magnetic moment of the loop is:

\[ \mu = \frac{1}{2} q \omega r^2 \]

This matches the given options in the question.