The self inductance of a toroid is :-

Options

A. `(mu _(0)N^(2)r^(2))/(2 R_(m))`
B. `(mu _(0)N^(2) pi r)/(2 R _(m))`
C. `(mu _(0)N^(2)r)/(2R _(m))`
D. `(mu _(0)N^(2)r pi)/(r_(m))`

Answer

To find the self-inductance of a toroid, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Geometry of the Toroid**:
- A toroid is a doughnut-shaped coil of wire. We denote the radius from the center of the toroid to the center of the wire as \( R \), and the radius of the wire itself as \( r \).

2. **Magnetic Field Inside the Toroid**:
- The magnetic field \( B \) inside a toroid carrying a current \( I \) is given by the formula:
\[ B = \frac{\mu_0 I N}{2 \pi R} \]
where \( \mu_0 \) is the permeability of free space and \( N \) is the number of turns of the wire.

3. **Calculating the Magnetic Flux**:
- The magnetic flux \( \Phi \) through one loop of the toroid can be expressed as:
\[ \Phi = B \cdot A \]
where \( A \) is the cross-sectional area of the toroid. For a circular cross-section, the area \( A \) can be calculated as:
\[ A = \pi r^2 \]
Therefore, the total magnetic flux through \( N \) turns is:
\[ \Phi_{\text{total}} = N \cdot B \cdot A = N \cdot \left(\frac{\mu_0 I N}{2 \pi R}\right) \cdot (\pi r^2) \]

4. **Relating Flux to Inductance**:
- The self-inductance \( L \) is defined by the relationship:
\[ \Phi_{\text{total}} = L \cdot I \]
Substituting the expression for total magnetic flux:
\[ N \cdot \left(\frac{\mu_0 I N}{2 \pi R}\right) \cdot (\pi r^2) = L \cdot I \]

5. **Solving for Self-Inductance \( L \)**:
- We can cancel \( I \) from both sides (assuming \( I \neq 0 \)):
\[ L = \frac{N^2 \mu_0 r^2}{2 \pi R} \]

6. **Final Expression for Self-Inductance**:
- Thus, the self-inductance of a toroid is given by:
\[ L = \frac{\mu_0 N^2 r^2}{2 \pi R} \]

### Summary:
The self-inductance of a toroid is:
\[ L = \frac{\mu_0 N^2 r^2}{2 \pi R} \]