A current I flows in a circular coil of radius r. If the coil is placed in a uniform magnetic field B with its plane parallel to the field, magnitude of the torque that acts on the coil is

Options

A. Zero
B. `2piriB`
C. `pir^(2)iB`
D. `2pir^(2)iB`

Answer

To find the magnitude of the torque acting on a circular coil of radius \( r \) carrying a current \( I \) when placed in a uniform magnetic field \( B \) with its plane parallel to the field, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Situation**:
- A circular coil is carrying a current \( I \).
- The coil is placed in a uniform magnetic field \( B \).
- The plane of the coil is parallel to the magnetic field.

2. **Identify the Torque Formula**:
- The torque \( \tau \) acting on a magnetic dipole in a magnetic field is given by the formula:
\[ \tau = \vec{m} \times \vec{B} \]
- This can also be expressed in terms of magnitude as:
\[ \tau = mB \sin \theta \]
- Here, \( m \) is the magnetic moment, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the magnetic moment and the magnetic field.

3. **Determine the Angle**:
- Since the plane of the coil is parallel to the magnetic field, the angle \( \theta \) between the magnetic moment \( \vec{m} \) and the magnetic field \( \vec{B} \) is \( 90^\circ \).
- Therefore, \( \sin 90^\circ = 1 \).

4. **Calculate the Magnetic Moment**:
- The magnetic moment \( m \) for a circular coil is given by:
\[ m = I \cdot A \]
- Where \( A \) is the area of the coil. For a circular coil of radius \( r \):
\[ A = \pi r^2 \]
- Thus, substituting for \( A \):
\[ m = I \cdot \pi r^2 \]

5. **Substitute into the Torque Formula**:
- Now substituting \( m \) back into the torque formula:
\[ \tau = mB \sin 90^\circ = (I \cdot \pi r^2) \cdot B \cdot 1 \]
- Therefore, the magnitude of the torque is:
\[ \tau = I \cdot \pi r^2 \cdot B \]

### Final Answer:
The magnitude of the torque acting on the coil is:
\[ \tau = I \cdot \pi r^2 \cdot B \]

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