A toroid having 200 turns carries a current of 1A. The average radius of the toroid is 10 cm. the magnetic field at any point in the open space inside the toroid is

Options

A. `4xx10^(-3)T`
B. zero
C. `0.5xx10^(-3)T`
D. `2xx10^(-3)T`

Answer

To find the magnetic field inside a toroid, we can use the formula for the magnetic field \( B \) at a distance \( r \) from the center of the toroid:

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

Where:
- \( \mu_0 \) is the permeability of free space, approximately \( 4 \pi \times 10^{-7} \, \text{T m/A} \)
- \( n \) is the number of turns per unit length
- \( I \) is the current flowing through the toroid
- \( r \) is the average radius of the toroid

### Step-by-Step Solution:

1. **Identify the given values:**
- Number of turns \( N = 200 \)
- Current \( I = 1 \, \text{A} \)
- Average radius \( r = 10 \, \text{cm} = 0.1 \, \text{m} \)

2. **Calculate the circumference of the toroid:**
\[ \text{Circumference} = 2 \pi r = 2 \pi (0.1) = 0.2 \pi \, \text{m} \]

3. **Calculate the number of turns per unit length \( n \):**
\[ n = \frac{N}{\text{Circumference}} = \frac{200}{0.2 \pi} = \frac{200}{0.628} \approx 318.31 \, \text{turns/m} \]

4. **Substitute the values into the magnetic field formula:**
\[ B = \mu_0 n I = (4 \pi \times 10^{-7}) \times (318.31) \times (1) \]

5. **Calculate the magnetic field \( B \):**
\[ B = 4 \pi \times 10^{-7} \times 318.31 \approx 4 \times 10^{-3} \, \text{T} \]

6. **Final Result:**
The magnetic field at any point in the open space inside the toroid is approximately \( 4 \times 10^{-3} \, \text{T} \).