A moving coil galvanometer has N number of turns in a coil of effective area. A it carries a current I. The magnetic field B is radial. The torque acting on the coil is

Options

A. `NA^(2)BI`
B. `NABI^(2)`
C. `N^(2)ABI`
D. NABI

Explanation

To find the torque acting on a moving coil galvanometer, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a coil with \( N \) turns, an effective area \( A \), and it carries a current \( I \). The magnetic field \( B \) is radial.

2. **Magnetic Dipole Moment**:
- The magnetic dipole moment \( \vec{M} \) of the coil can be defined as:
\[ \vec{M} = N \cdot I \cdot \vec{A} \]
- Here, \( \vec{A} \) is the area vector of the coil, which is perpendicular to the plane of the coil.

3. **Torque on the Coil**:
- The torque \( \vec{\tau} \) acting on a magnetic dipole in a magnetic field is given by the formula:
\[ \vec{\tau} = \vec{M} \times \vec{B} \]
- The magnitude of the torque can be expressed as:
\[ \tau = M \cdot B \cdot \sin(\theta) \]
- Where \( \theta \) is the angle between the magnetic dipole moment \( \vec{M} \) and the magnetic field \( \vec{B} \).

4. **Angle Consideration**:
- In this scenario, since the magnetic field \( B \) is radial and the area vector \( \vec{A} \) is perpendicular to the plane of the coil, the angle \( \theta \) between \( \vec{M} \) and \( \vec{B} \) is \( 90^\circ \). Thus, \( \sin(90^\circ) = 1 \).

5. **Substituting Values**:
- Therefore, the torque simplifies to:
\[ \tau = M \cdot B \]
- Substituting the expression for \( M \):
\[ \tau = (N \cdot I \cdot A) \cdot B \]

6. **Final Expression**:
- Thus, the torque acting on the coil is:
\[ \tau = N \cdot I \cdot A \cdot B \]

### Final Answer:
The torque acting on the coil is \( \tau = N I A B \).

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