A circular coil of radius r having number of turns n and carrying a current A produces magnetic induction at its centre of magnitude B. B can be doubled by

Options

A. keeping the number of turns n and changing the current to A/2
B. changing the number of turns to n / 2 and keeping the current at A
C. simultaneously changing the number of turns and current to 2n and 2A
D. keeping the current at A and changing the number of turns to 2n.

Answer

To solve the problem of how to double the magnetic induction \( B \) at the center of a circular coil, we start with the formula for magnetic induction due to a circular coil:

### Step 1: Write the formula for magnetic induction
The magnetic induction \( B \) at the center of a circular coil is given by the formula:
\[ B = \frac{n \mu_0 I}{2r} \]
where:
- \( n \) = number of turns in the coil
- \( \mu_0 \) = permeability of free space
- \( I \) = current flowing through the coil
- \( r \) = radius of the coil

### Step 2: Analyze the conditions to double \( B \)
To double the magnetic induction, we need to find a condition such that:
\[ B' = 2B \]
Substituting the expression for \( B \):
\[ B' = \frac{n' \mu_0 I'}{2r} \]
We want:
\[ \frac{n' \mu_0 I'}{2r} = 2 \left( \frac{n \mu_0 I}{2r} \right) \]
This simplifies to:
\[ n' I' = 4n I \]

### Step 3: Evaluate the options
1. **Keeping \( n \) constant and changing \( I \) to \( \frac{I}{2} \)**:
\[ n' = n, \quad I' = \frac{I}{2} \]
\[ n' I' = n \cdot \frac{I}{2} = \frac{n I}{2} \quad \text{(not doubled)} \]

2. **Changing \( n \) to \( \frac{n}{2} \) and keeping \( I \) constant**:
\[ n' = \frac{n}{2}, \quad I' = I \]
\[ n' I' = \frac{n}{2} \cdot I = \frac{n I}{2} \quad \text{(not doubled)} \]

3. **Simultaneously changing \( n \) to \( 2n \) and \( I \) to \( 2I \)**:
\[ n' = 2n, \quad I' = 2I \]
\[ n' I' = 2n \cdot 2I = 4n I \quad \text{(this quadruples \( B \))} \]

4. **Keeping \( I \) constant and changing \( n \) to \( 2n \)**:
\[ n' = 2n, \quad I' = I \]
\[ n' I' = 2n \cdot I = 2n I \quad \text{(this doubles \( B \))} \]

### Conclusion
The correct condition to double the magnetic induction \( B \) is to keep the current \( I \) constant and change the number of turns \( n \) to \( 2n \).

Thus, the answer is:
**Option 4: Keeping the current \( I \) constant and changing the number of turns to \( 2n \).**

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