A length of wire carries a steady current. It is first bent to form a circular coil of one turn. The same length is now bent more sharply to give a loop of two turns of smaller radius. The magentic field at the centre caused by the same current now will be

Options

A. A quarter of its first is
B. unaltered
C. four times of ots first value
D. two times of its first value

Answer

To solve the problem, we need to analyze the magnetic field produced by the wire when it is bent into different shapes while carrying the same current.

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a wire of a certain length that carries a steady current \( I \).
- First, it is bent into a circular coil of one turn.
- Then, it is bent into a loop of two turns with a smaller radius.

2. **Length of the Wire**:
- Let the radius of the first coil be \( r_1 \). The length of the wire used for the first coil is given by:
\[ L = 2\pi r_1 \]
- For the second configuration, let the radius be \( r_2 \). Since there are two turns, the length of the wire is:
\[ L = 2 \times 2\pi r_2 = 4\pi r_2 \]
- Since the length of the wire remains constant, we can equate the two expressions:
\[ 2\pi r_1 = 4\pi r_2 \]
- Simplifying this gives:
\[ r_1 = 2r_2 \]

3. **Magnetic Field Calculation**:
- The magnetic field \( B \) at the center of a circular loop carrying current \( I \) is given by the formula:
\[ B = \frac{\mu_0 I}{2r} \]
- For the first coil (one turn), the magnetic field \( B_1 \) is:
\[ B_1 = \frac{\mu_0 I}{2r_1} \]
- For the second coil (two turns), the magnetic field \( B_2 \) is:
\[ B_2 = \frac{2 \mu_0 I}{2r_2} = \frac{\mu_0 I}{r_2} \]

4. **Relating \( B_1 \) and \( B_2 \)**:
- Now, substituting \( r_1 = 2r_2 \) into the expression for \( B_1 \):
\[ B_1 = \frac{\mu_0 I}{2(2r_2)} = \frac{\mu_0 I}{4r_2} \]
- Now, we can find the ratio of the two magnetic fields:
\[ \frac{B_2}{B_1} = \frac{\frac{\mu_0 I}{r_2}}{\frac{\mu_0 I}{4r_2}} = \frac{4}{1} = 4 \]

5. **Conclusion**:
- Therefore, the magnetic field at the center caused by the two-turn loop is four times that of the single-turn coil:
\[ B_2 = 4B_1 \]

### Final Answer:
The magnetic field at the center caused by the two-turn loop is **4 times** the magnetic field of the single-turn coil.

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