A circular coil of radius R and a current `I`, which can rotate about a fixed axis passing through its diameter is initially placed such that its plane lies along magnetic field B kinetic energy of loop when it rotates through an angle `90^(@)` is : (Assume that `I` remains constant)

Options

A. `pi R^(2) BI`
B. `(pi R^(2) BI)/(2)`
C. `2 piR^(2)BI`
D. `(3)/(2) pi R^(2) I`

Explanation

To solve the problem, we need to determine the kinetic energy of a circular coil when it rotates through an angle of \(90^\circ\) in a magnetic field \(B\). Here are the steps to arrive at the solution:

### Step 1: Understand the Initial Setup
The circular coil has a radius \(R\) and carries a current \(I\). It is initially positioned such that its plane is parallel to the magnetic field \(B\).

### Step 2: Determine the Magnetic Moment
The magnetic moment \(M\) of the coil can be calculated using the formula:
\[ M = I \cdot A \]
where \(A\) is the area of the coil. The area \(A\) of a circular coil is given by:
\[ A = \pi R^2 \]
Thus, the magnetic moment becomes:
\[ M = I \cdot \pi R^2 \]

### Step 3: Calculate the Change in Potential Energy
When the coil rotates through an angle of \(90^\circ\), the angle between the magnetic moment and the magnetic field changes. Initially, the angle \(\theta\) is \(0^\circ\) (since the plane of the coil is parallel to the magnetic field), and after rotation, \(\theta\) becomes \(90^\circ\).

The potential energy \(U\) in a magnetic field is given by:
\[ U = -M \cdot B \cdot \cos(\theta) \]
Initially, when \(\theta = 0^\circ\):
\[ U_{\text{initial}} = -M \cdot B \cdot \cos(0) = -M \cdot B \]
After rotating to \(90^\circ\):
\[ U_{\text{final}} = -M \cdot B \cdot \cos(90) = 0 \]

### Step 4: Calculate the Change in Potential Energy
The change in potential energy \(\Delta U\) as the coil rotates from \(0^\circ\) to \(90^\circ\) is:
\[ \Delta U = U_{\text{final}} - U_{\text{initial}} = 0 - (-M \cdot B) = M \cdot B \]

### Step 5: Relate Change in Potential Energy to Kinetic Energy
According to the principle of conservation of energy, the loss in potential energy is equal to the gain in kinetic energy \(K\):
\[ K = \Delta U = M \cdot B \]

### Step 6: Substitute the Expression for Magnetic Moment
Substituting the expression for the magnetic moment \(M\):
\[ K = (I \cdot \pi R^2) \cdot B \]

### Final Answer
Thus, the kinetic energy of the loop when it rotates through an angle of \(90^\circ\) is:
\[ K = I \cdot \pi R^2 \cdot B \]