A current of 2 A is made to flow through a coil which has only one turn. The magnetic field produced at the centre is`4pi xx 10^(-6) "Wb"//m^(2).` the radius of the coil is

Options

A. `0.0001m`
B. `0.01 m`
C. `0.1 m`
D. `0.001 m`

Answer

To find the radius of the coil given the magnetic field at its center, we can use the formula for the magnetic field produced by a circular coil:

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

Where:
- \( B \) = magnetic field at the center of the coil (in Weber/m²)
- \( \mu_0 \) = permeability of free space (\( 4\pi \times 10^{-7} \, \text{T m/A} \))
- \( I \) = current flowing through the coil (in Amperes)
- \( N \) = number of turns in the coil
- \( R \) = radius of the coil (in meters)

Given:
- \( B = 4\pi \times 10^{-6} \, \text{Wb/m}^2 \)
- \( I = 2 \, \text{A} \)
- \( N = 1 \) (since the coil has only one turn)

### Step 1: Substitute the known values into the formula

We can rewrite the formula for our specific case (with \( N = 1 \)):

\[ B = \frac{\mu_0 I}{2R} \]

Substituting the known values:

\[ 4\pi \times 10^{-6} = \frac{(4\pi \times 10^{-7}) \times 2}{2R} \]

### Step 2: Simplify the equation

The \( 2 \) in the numerator and the \( 2 \) in the denominator cancel out:

\[ 4\pi \times 10^{-6} = \frac{4\pi \times 10^{-7}}{R} \]

### Step 3: Rearrange to solve for \( R \)

Now, we can rearrange the equation to solve for \( R \):

\[ R = \frac{4\pi \times 10^{-7}}{4\pi \times 10^{-6}} \]

### Step 4: Simplify further

The \( 4\pi \) terms cancel out:

\[ R = \frac{10^{-7}}{10^{-6}} \]

Using the property of exponents:

\[ R = 10^{-7 + 6} = 10^{-1} \]

### Step 5: Convert to meters

Thus, we find:

\[ R = 0.1 \, \text{m} \]

### Final Answer

The radius of the coil is \( 0.1 \, \text{m} \) or \( 10 \, \text{cm} \).

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