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 at any point in the open space inside a toroid, we can use the formula for the magnetic field inside a toroid, which is given by:

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

where:
- \( B \) is the magnetic field,
- \( \mu_0 \) is the permeability of free space (\( 4 \pi \times 10^{-7} \, \text{T m/A} \)),
- \( N \) is the total number of turns,
- \( I \) is the current in amperes,
- \( r \) is the average radius of the toroid in meters.

### Step-by-Step Solution:

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

2. **Convert the radius to meters:**
- Since \( r = 10 \, \text{cm} \), we convert it to meters:
\[ r = \frac{10}{100} = 0.1 \, \text{m} \]

3. **Substitute the values into the formula:**
\[ B = \frac{\mu_0 N I}{2 \pi r} \]
Substituting the known values:
\[ B = \frac{(4 \pi \times 10^{-7} \, \text{T m/A}) \times 200 \times 1}{2 \pi \times 0.1} \]

4. **Simplify the expression:**
- The \( \pi \) in the numerator and denominator cancels out:
\[ B = \frac{(4 \times 10^{-7} \times 200)}{2 \times 0.1} \]
- Calculate the denominator:
\[ 2 \times 0.1 = 0.2 \]
- Now substitute this back into the equation:
\[ B = \frac{800 \times 10^{-7}}{0.2} \]

5. **Calculate the magnetic field:**
\[ B = 800 \times 10^{-7} \div 0.2 = 800 \times 10^{-7} \times 5 = 4000 \times 10^{-7} = 4 \times 10^{-4} \, \text{T} \]

6. **Final result:**
\[ B = 4 \times 10^{-3} \, \text{T} \text{ or } 4 \, \text{mT} \]

### Final Answer:
The magnetic field at any point in the open space inside the toroid is \( 4 \, \text{mT} \).