For toroid of mean radius 20 cm, turns = 240 and a current of 2A. What is the value of magnetic field within toroid ?

Options

A. `4.8xx10^(-4)T`
B. `2.4xx10^(-4)T`
C. `1.2xx10^(-4)T`
D. Zero

Answer

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

\[ B = \mu_0 \cdot n \cdot I \]

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

### Step-by-Step Solution:

1. **Identify Given Values:**
- Mean radius of the toroid, \( R = 20 \, \text{cm} = 0.2 \, \text{m} \)
- Number of turns, \( N = 240 \)
- Current, \( I = 2 \, \text{A} \)

2. **Calculate the Circumference of the Toroid:**
The length \( L \) of the toroid can be calculated using the formula for the circumference of a circle:
\[ L = 2\pi R = 2\pi \times 0.2 \, \text{m} = 0.4\pi \, \text{m} \]

3. **Calculate the Number of Turns per Unit Length (n):**
The number of turns per unit length \( n \) is given by:
\[ n = \frac{N}{L} = \frac{240}{0.4\pi} \]
Simplifying this gives:
\[ n = \frac{240}{0.4 \times 3.14} \approx \frac{240}{1.256} \approx 191.07 \, \text{turns/m} \]

4. **Substitute Values into the Magnetic Field Formula:**
Now substitute \( n \) and \( I \) into the magnetic field formula:
\[ B = \mu_0 \cdot n \cdot I = (4\pi \times 10^{-7}) \cdot (191.07) \cdot (2) \]

5. **Calculate the Magnetic Field:**
\[ B = (4\pi \times 10^{-7}) \cdot 191.07 \cdot 2 \approx (4 \times 3.14 \times 10^{-7}) \cdot 191.07 \cdot 2 \]
\[ B \approx (12.56 \times 10^{-7}) \cdot 191.07 \cdot 2 \approx 4.8 \times 10^{-4} \, \text{T} \]

6. **Final Result:**
The magnetic field within the toroid is approximately:
\[ B \approx 4.8 \times 10^{-4} \, \text{T} \]