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 Ampere's circuital law. The formula for the magnetic field \( B \) inside a toroid is given by:

\[ B = \frac{\mu_0 N I}{2 \pi r} \]

Where:
- \( B \) is the magnetic field inside the toroid,
- \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \),
- \( N \) is the total number of turns,
- \( I \) is the current flowing through the wire,
- \( r \) is the radius of the toroid.

However, in this case, we are given the number of turns per unit length \( n \) and the current \( I \).

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. **Calculate the total number of turns \( N \):**
Since we are not given the radius, we can express the magnetic field in terms of \( n \):
\[ N = n \cdot L \]
where \( L \) is the length of the toroid. We can use \( n \) directly in the formula.

3. **Substituting into the formula:**
The formula for the magnetic field inside the toroid can be simplified using \( n \):
\[ B = \mu_0 n I \]

4. **Substituting the values:**
\[ B = (4\pi \times 10^{-7}) \times (1000) \times \left(\frac{1}{4\pi}\right) \]

5. **Simplifying the expression:**
\[ B = (4\pi \times 10^{-7}) \times 1000 \times \frac{1}{4\pi} = 10^{-4} \, \text{T} \]

6. **Convert to Weber/m²:**
Since \( 1 \, \text{T} = 1 \, \text{Wb/m}^2 \), we have:
\[ B = 10^{-4} \, \text{Wb/m}^2 \]

Thus, the magnetic field produced inside the toroid is \( 10^{-4} \, \text{Wb/m}^2 \).