A moving coil galvanometer has `N` numbr 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)B^(2)I`
B. `NABI^92)`
C. `N^(2)ABI`
D. NABI

Answer

To solve the problem of finding the torque acting on a moving coil galvanometer, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Torque Formula**:
The torque (\(\tau\)) acting on a coil in a magnetic field can be expressed as:
\[ \tau = \mu \times B \]
where \(\mu\) is the magnetic moment of the coil and \(B\) is the magnetic field.

2. **Define the Magnetic Moment**:
The magnetic moment (\(\mu\)) for a coil with \(N\) turns, carrying a current \(I\), and having an effective area \(A\) is given by:
\[ \mu = N \cdot I \cdot A \]

3. **Substitute the Magnetic Moment into the Torque Formula**:
Replacing \(\mu\) in the torque formula, we get:
\[ \tau = (N \cdot I \cdot A) \times B \]

4. **Consider the Angle Between \(\mu\) and \(B\)**:
The torque can also be expressed in terms of the angle \(\theta\) between the magnetic moment and the magnetic field:
\[ \tau = \mu B \sin(\theta) \]
If the magnetic field \(B\) is radial and the angle \(\theta\) is \(90^\circ\) (which means the plane of the coil is perpendicular to the magnetic field), then \(\sin(90^\circ) = 1\).

5. **Final Expression for Torque**:
Thus, substituting \(\theta = 90^\circ\) into the torque equation gives:
\[ \tau = N \cdot I \cdot A \cdot B \]

### Conclusion:
The torque acting on the coil of the moving coil galvanometer is given by:
\[ \tau = N \cdot I \cdot A \cdot B \]