The electric current in a circular coil of two turns produced a magnetic induction of 0.2 T at its centre. The coil is unwound and then rewound into a circular coil of four turns. If same current flows in the coil, the magnetic induction at the centre of the coil now is

Options

A. 0.2T
B. 0.4T
C. 0.6T
D. 0.8T

Explanation

To solve the problem, we need to understand the relationship between the magnetic induction (B) at the center of a circular coil, the number of turns (n), and the radius (r) of the coil.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- For the first coil:
- Number of turns, \( n_1 = 2 \)
- Magnetic induction, \( B_1 = 0.2 \, \text{T} \)
- For the second coil:
- Number of turns, \( n_2 = 4 \)

2. **Understand the Formula:**
The magnetic induction at the center of a circular coil can be expressed as:
\[ B \propto \frac{n}{r} \]
where \( n \) is the number of turns and \( r \) is the radius of the coil.

3. **Relate the Two Cases:**
Since the coil is unwound and rewound, the length of wire used remains constant. The length of wire for the first coil is:
\[ L_1 = n_1 \cdot 2\pi r_1 = 2 \cdot 2\pi r_1 = 4\pi r_1 \]
For the second coil:
\[ L_2 = n_2 \cdot 2\pi r_2 = 4 \cdot 2\pi r_2 = 8\pi r_2 \]
Setting \( L_1 = L_2 \):
\[ 4\pi r_1 = 8\pi r_2 \]
Simplifying gives:
\[ r_1 = 2r_2 \]

4. **Substituting into the Magnetic Induction Formula:**
Using the relationship \( B \propto \frac{n}{r} \), we can set up the ratio of the magnetic inductions:
\[ \frac{B_1}{B_2} = \frac{n_1 \cdot r_2}{n_2 \cdot r_1} \]
Substituting \( n_1 = 2 \), \( n_2 = 4 \), and \( r_1 = 2r_2 \):
\[ \frac{B_1}{B_2} = \frac{2 \cdot r_2}{4 \cdot (2r_2)} = \frac{2}{8} = \frac{1}{4} \]

5. **Finding \( B_2 \):**
Rearranging gives:
\[ B_2 = 4B_1 \]
Substituting \( B_1 = 0.2 \, \text{T} \):
\[ B_2 = 4 \times 0.2 = 0.8 \, \text{T} \]

### Final Answer:
The magnetic induction at the center of the coil with 4 turns is \( B_2 = 0.8 \, \text{T} \).

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