A circular coil having N turns and radius r carries a current I. It is held in the XZ plane in a magnetic field `Bhati`. The torque on the coil due to the magnetic field is :

Options

A. `B pi r^(2) IN`
B. Zero
C. `(Br^(2)I)/(piN)`
D. `(B pi r^(2) I)/N`

Explanation

To find the torque on a circular coil due to a magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a circular coil with \( N \) turns and radius \( r \).
- The coil carries a current \( I \) and is placed in the XZ plane.
- The magnetic field is directed along the \( \hat{i} \) (x-direction).

2. **Direction of Magnetic Moment**:
- The magnetic moment \( \vec{m} \) of the coil is given by the formula:
\[ \vec{m} = N \cdot I \cdot A \]
where \( A \) is the area of the coil.
- The area \( A \) of a circular coil is calculated as:
\[ A = \pi r^2 \]
- Therefore, the magnetic moment becomes:
\[ \vec{m} = N \cdot I \cdot \pi r^2 \]

3. **Torque Calculation**:
- The torque \( \vec{\tau} \) on the coil in a magnetic field \( \vec{B} \) is given by the cross product:
\[ \vec{\tau} = \vec{m} \times \vec{B} \]
- The magnitude of the torque can also be expressed as:
\[ \tau = mB \sin \theta \]
where \( \theta \) is the angle between the magnetic moment and the magnetic field.

4. **Finding the Angle**:
- In this case, since the coil is in the XZ plane and the magnetic field is along the x-direction, the angle \( \theta \) between \( \vec{m} \) and \( \vec{B} \) is \( 90^\circ \).
- Therefore, \( \sin 90^\circ = 1 \).

5. **Substituting Values**:
- Now substituting the values into the torque equation:
\[ \tau = mB \cdot 1 = mB \]
- Substituting \( m = N \cdot I \cdot \pi r^2 \):
\[ \tau = (N \cdot I \cdot \pi r^2) \cdot B \]

6. **Final Expression**:
- Thus, the torque on the coil due to the magnetic field is:
\[ \tau = N \cdot I \cdot \pi r^2 \cdot B \]

### Conclusion:
The torque on the coil due to the magnetic field is given by:
\[ \tau = N \cdot I \cdot \pi r^2 \cdot B \]