A straight wire of diameter `0.5 mm` carrying a current of `1 A` is replaced by another wire of `1 mm` diameter carrying the same current. The strength of magnetic field far away is

Options

A. Twice the earlier value
B. Half of the earlier value
C. Quarter of its earlier value
D. Unchanged

Answer

To solve the problem, we need to determine the strength of the magnetic field produced by a straight wire carrying a current. The magnetic field (B) created by a long straight wire at a distance (r) from the wire is given by the formula:

\[ B = \frac{\mu_0 I}{2 \pi r} \]

where:
- \( B \) is the magnetic field strength,
- \( \mu_0 \) is the permeability of free space (approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \)),
- \( I \) is the current in amperes,
- \( r \) is the distance from the wire in meters.

### Step-by-Step Solution:

1. **Identify the Parameters**:
- Current \( I = 1 \, \text{A} \)
- The diameter of the wire does not affect the magnetic field strength at a distance far away from the wire.

2. **Understand the Dependence of Magnetic Field**:
- The magnetic field strength \( B \) is directly proportional to the current \( I \) and inversely proportional to the distance \( r \).
- Importantly, the magnetic field strength does not depend on the diameter of the wire when calculating the field at a distance far away.

3. **Calculate the Magnetic Field**:
- Since both wires carry the same current of \( 1 \, \text{A} \) and we are interested in the magnetic field at a distance far away from the wire, we can conclude that the magnetic field strength will be the same for both wires.

4. **Conclusion**:
- The strength of the magnetic field far away from both wires is unchanged and is given by the formula:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
- Since the current \( I \) remains constant, the magnetic field strength will remain the same regardless of the wire's diameter.

### Final Answer:
The strength of the magnetic field far away from both wires is unchanged.