A toroid having average diameter 2.5 m, number A turns 400 , current=2A and magnetic field has 10T what will be induced magnetic field (in amp/m)

Options

A. `(10^(5))/(4pi)`
B. `(10^(8))/(4pi)`
C. `(10^(8))/(2pi)`
D. `(10^(2))/(2pi)`

Explanation

To solve the problem step-by-step, we will use the relevant formulas and given data.

### Step 1: Identify the given data
- Average diameter of the toroid, \( D = 2.5 \, \text{m} \)
- Number of turns, \( N = 400 \)
- Current, \( I = 2 \, \text{A} \)
- Magnetic field, \( B = 10 \, \text{T} \)

### Step 2: Calculate the radius of the toroid
The radius \( r \) is half of the diameter:
\[ r = \frac{D}{2} = \frac{2.5}{2} = 1.25 \, \text{m} \]

### Step 3: Use the formula for the magnetic field in a toroid
The magnetic field \( B \) inside a toroid is given by:
\[ B = \mu_0 \cdot H \]
where \( H \) is the magnetic field intensity given by:
\[ H = \frac{N \cdot I}{2 \pi r} \]
Substituting for \( H \):
\[ B = \mu_0 \cdot \frac{N \cdot I}{2 \pi r} \]

### Step 4: Rearranging the equation to find the induced magnetic field \( I \)
We can rearrange the equation to isolate the induced magnetic field \( I \):
\[ I = \frac{B \cdot 2 \pi r}{\mu_0 \cdot N} \]

### Step 5: Substitute the known values
We know:
- \( B = 10 \, \text{T} \)
- \( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \)
- \( N = 400 \)
- \( r = 1.25 \, \text{m} \)

Substituting these values into the equation:
\[ I = \frac{10 \cdot 2 \pi \cdot 1.25}{4 \pi \times 10^{-7} \cdot 400} \]

### Step 6: Simplifying the equation
\[ I = \frac{10 \cdot 2 \cdot 1.25}{4 \times 10^{-7} \cdot 400} \]
\[ I = \frac{25}{4 \times 10^{-7} \cdot 400} \]
\[ I = \frac{25}{1.6 \times 10^{-4}} = \frac{25 \times 10^4}{1.6} \]
\[ I = \frac{250000}{1.6} = 156250 \]

### Step 7: Final answer
Thus, the induced magnetic field \( I \) is:
\[ I = 156250 \, \text{A/m} \]