In a toroid the number of tuns per unit length is 1000 and current through it is `1/(4pi)` ampere. The magnetic field produced inside (in `"weber"//m^(2)`) will be

Options

A. `10^(-2)`
B. `10^(-3)`
C. `10^(-4)`
D. `10^(-7)`

Answer

To find the magnetic field produced inside a toroid, we can use the formula:

\[ B = \mu_0 n I \]

where:
- \( B \) is the magnetic field,
- \( \mu_0 \) is the permeability of free space, which is approximately \( 4\pi \times 10^{-7} \, \text{Wb/m}^2 \),
- \( n \) is the number of turns per unit length,
- \( I \) is the current flowing through the toroid.

### Step-by-Step Solution:

1. **Identify the given values:**
- Number of turns per unit length, \( n = 1000 \, \text{turns/m} \)
- Current, \( I = \frac{1}{4\pi} \, \text{A} \)

2. **Substitute the values into the formula:**
\[ B = \mu_0 n I \]
Substituting \( \mu_0 = 4\pi \times 10^{-7} \, \text{Wb/m}^2 \), \( n = 1000 \, \text{turns/m} \), and \( I = \frac{1}{4\pi} \):
\[ B = (4\pi \times 10^{-7}) \times 1000 \times \left(\frac{1}{4\pi}\right) \]

3. **Simplify the expression:**
- The \( 4\pi \) in the numerator and denominator cancels out:
\[ B = (10^{-7} \times 1000) = 10^{-7} \times 10^3 \]
- This simplifies to:
\[ B = 10^{-4} \, \text{Wb/m}^2 \]

4. **Final Result:**
The magnetic field produced inside the toroid is:
\[ B = 10^{-4} \, \text{Wb/m}^2 \]