The current in the windings on a toroid is 2.0A. There are 400 turns and the mean circumferential length is 40cm. If the inside magnetic field is 1.0T, the relative permeability is near to

Options

A. 100
B. 200
C. 300
D. 400

Explanation

To find the relative permeability (\( \mu_r \)) of the material in the toroid, we can use the following steps:

### Step 1: Understand the formula for magnetic field in a toroid
The magnetic field (\( B \)) inside a toroid is given by the formula:
\[ B = \mu H \]
where \( \mu \) is the permeability of the material and \( H \) is the magnetic field strength.

### Step 2: Calculate the magnetic field strength (\( H \))
The magnetic field strength (\( H \)) in a toroid can be calculated using the formula:
\[ H = \frac{NI}{l} \]
where:
- \( N \) = number of turns (400 turns)
- \( I \) = current (2.0 A)
- \( l \) = mean circumferential length (40 cm = 0.4 m)

Substituting the values:
\[ H = \frac{400 \times 2.0}{0.4} = \frac{800}{0.4} = 2000 \, \text{A/m} \]

### Step 3: Relate \( B \), \( \mu \), and \( H \)
We know that:
\[ B = \mu H \]
We can express \( \mu \) as:
\[ \mu = \frac{B}{H} \]

### Step 4: Substitute the values of \( B \) and \( H \)
Given \( B = 1.0 \, \text{T} \) and \( H = 2000 \, \text{A/m} \):
\[ \mu = \frac{1.0}{2000} = 0.0005 \, \text{T m/A} \]

### Step 5: Calculate the permeability of free space (\( \mu_0 \))
The permeability of free space (\( \mu_0 \)) is approximately:
\[ \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \approx 1.2566 \times 10^{-6} \, \text{T m/A} \]

### Step 6: Calculate the relative permeability (\( \mu_r \))
The relative permeability is given by:
\[ \mu_r = \frac{\mu}{\mu_0} \]
Substituting the values:
\[ \mu_r = \frac{0.0005}{1.2566 \times 10^{-6}} \approx 398.1 \]

### Step 7: Round the result
Rounding \( \mu_r \) gives us approximately:
\[ \mu_r \approx 400 \]

### Final Answer
Thus, the relative permeability is approximately \( 400 \).

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