The magnetic field of a given length of wire carrying a current of a single turn circular coil at cebtre is B, then its value for two turns for the same wire when same current passing through it is

Options

A. `B/4`
B. `B/2`
C. `2B`
D. `4B`

Answer

To find the magnetic field at the center of a circular coil when the number of turns is doubled while keeping the same current, we can follow these steps:

### Step 1: Understand the formula for magnetic field
The magnetic field \( B \) at the center of a circular coil is given by the formula:
\[ B = \frac{\mu_0 n I}{2r} \]
where:
- \( B \) is the magnetic field,
- \( \mu_0 \) is the permeability of free space,
- \( n \) is the number of turns per unit length,
- \( I \) is the current through the wire,
- \( r \) is the radius of the coil.

### Step 2: Analyze the initial conditions
For a single turn circular coil, we have:
- Number of turns \( n = 1 \)
- Current \( I \) is constant
- Radius \( r \)

Thus, the magnetic field at the center is:
\[ B = \frac{\mu_0 (1) I}{2r} = \frac{\mu_0 I}{2r} \]

### Step 3: Modify for two turns
When we double the number of turns to \( n = 2 \), we need to determine the new radius. The length of the wire remains the same, and the circumference of the coil for two turns will be:
\[ L = 2 \times 2\pi r' \quad \text{(for two turns)} \]
where \( r' \) is the new radius.

Since the length of the wire is constant, we have:
\[ L = 2\pi r \quad \text{(for one turn)} \]
Setting the two lengths equal gives:
\[ 2\pi r' = \frac{L}{2} \quad \Rightarrow \quad r' = \frac{L}{4\pi} \]

### Step 4: Substitute into the magnetic field formula
Now substituting \( n = 2 \) and \( r = r' \) into the magnetic field formula:
\[ B' = \frac{\mu_0 (2) I}{2r'} = \frac{\mu_0 (2) I}{2 \left(\frac{L}{4\pi}\right)} = \frac{\mu_0 (2) I \cdot 4\pi}{2L} = \frac{4\mu_0 I}{L} \]

### Step 5: Relate the new magnetic field to the original
Since the original magnetic field was:
\[ B = \frac{\mu_0 I}{2r} \]
and we found that \( B' = 4B \), we can conclude that:
\[ B' = 4B \]

### Final Answer
Thus, the value of the magnetic field for two turns of the same wire with the same current is:
\[ B' = 4B \]

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