A toroid of 4000 turns has outer radius of 26 cm and inner radius of 25 cm. If the current in the wire is 10 A. Calculate the magnetic field of the toroid.

Answer

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

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

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.

### Step 1: Calculate the average radius \( r \)

The average radius \( r \) can be calculated as the mean of the inner radius \( r_{inner} \) and outer radius \( r_{outer} \):

\[ r = \frac{r_{inner} + r_{outer}}{2} \]

Given:
- \( r_{inner} = 25 \, \text{cm} = 0.25 \, \text{m} \)
- \( r_{outer} = 26 \, \text{cm} = 0.26 \, \text{m} \)

Calculating \( r \):

\[ r = \frac{0.25 + 0.26}{2} = \frac{0.51}{2} = 0.255 \, \text{m} \]

### Step 2: Substitute values into the magnetic field formula

Now we can substitute the values into the magnetic field formula. We know:
- \( N = 4000 \) turns,
- \( I = 10 \, \text{A} \),
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \).

Substituting these values into the formula:

\[ B = \frac{(4\pi \times 10^{-7}) \times 4000 \times 10}{2 \pi \times 0.255} \]

### Step 3: Simplify the expression

We can simplify the expression:

\[ B = \frac{4 \times 4000 \times 10 \times 10^{-7}}{2 \times 0.255} \]

Calculating the numerator:

\[ 4 \times 4000 \times 10 = 160000 \]

Now substituting back:

\[ B = \frac{160000 \times 10^{-7}}{2 \times 0.255} \]

Calculating the denominator:

\[ 2 \times 0.255 = 0.51 \]

So now we have:

\[ B = \frac{160000 \times 10^{-7}}{0.51} \]

### Step 4: Final calculation

Calculating \( B \):

\[ B = \frac{160000 \times 10^{-7}}{0.51} \approx 3.137 \times 10^{-2} \, \text{T} \]

### Final Answer

The magnetic field \( B \) in the toroid is approximately:

\[ B \approx 3.137 \times 10^{-2} \, \text{T} \]