A small coil of `N` turns has area `A` and a current `I` flows through it. The magnetic dipole moment of this coil will be

Options

A. `iNA`
B. `i^(2)NA`
C. `iN^(2)A`
D. `IN//A`

Answer

To find the magnetic dipole moment of a small coil with `N` turns, area `A`, and current `I` flowing through it, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Magnetic Dipole Moment**:
The magnetic dipole moment (μ) of a single loop of wire carrying current is defined as the product of the current (I) flowing through the loop and the area (A) of the loop.

\[ \mu = I \times A \]

2. **Extend to Multiple Turns**:
If the coil has `N` turns, each turn contributes to the total magnetic dipole moment. Therefore, the total magnetic dipole moment for `N` turns is simply `N` times the magnetic dipole moment of a single turn.

\[ \mu_{\text{total}} = N \times (I \times A) \]

3. **Final Expression**:
Thus, the expression for the magnetic dipole moment of the coil with `N` turns is:

\[ \mu_{\text{total}} = NIA \]

### Conclusion:
The magnetic dipole moment of the coil is given by:

\[ \mu = NIA \]