Question

The magnetic field at a point \( P \) at a distance of 2 cm from a long straight wire of diameter 0.5 mm carrying a current of 1 A is \( B \). If the diameter of the wire is doubled without changing the current, the magnetic field at the same point \( P \) is:

• A. \( 2B \)
• B. \( \frac{B}{2} \)
• C. \( \frac{3B}{4} \)
• D. \( B \)

Hint:

Magnetic field outside a long straight wire depends on distance from the wire center and current, not on the wire's diameter.

Answer 1

The Correct Option is D

The magnetic field \( B \) at a distance \( r \) from a long straight current-carrying wire is given by Ampère's law:

\[ B = \frac{\mu_0 I}{2\pi r} \]
where
- \( \mu_0 \) is the permeability of free space,
- \( I \) is the current in the wire,
- \( r \) is the perpendicular distance from the wire.

The diameter of the wire affects the thickness of the conductor but does not affect the magnetic field outside the wire at the point \( P \), as long as \( r \) is measured from the center of the wire and the current \( I \) remains the same.

Here, \( r = 2\, \text{cm} \), which is much larger than the wire radius (initially 0.25 mm, doubled to 0.5 mm). So, the magnetic field at \( P \) depends only on \( r \) and \( I \), both unchanged.

Therefore, the magnetic field remains:

\[ B = \text{same value} \]