Question

A long straight wire of radius \( a \) carries a steady current \( I \). The current is uniformly distributed across its cross-section. The ratio of the magnetic field at \( a/2 \) and \( 2a \) from the axis of the wire is:

• A. \( 1:4 \)
• B. \( 4:1 \)
• C. \( 1:1 \)
• D. \( 3:4 \)

Hint:

Magnetic field inside a current-carrying wire varies linearly, while outside it follows an inverse relation with distance.

Answer 1

The Correct Option is C

Step 1: {Using Ampere’s Circuital Law}
The magnetic field due to a straight wire is given by: \[ B = \frac{\mu_0 I}{2\pi r} \] Step 2: {Finding the Magnetic Field at \( a/2 \) and \( 2a \)}
Magnetic field at \( r = a/2 \): \[ B_{a/2} = \frac{\mu_0 I}{2\pi (a/2)} \] \[ B_{a/2} = \frac{\mu_0 I}{\pi a} \] Magnetic field at \( r = 2a \): \[ B_{2a} = \frac{\mu_0 I}{2\pi (2a)} \] \[ B_{2a} = \frac{\mu_0 I}{4\pi a} \] Step 3: {Calculating the Ratio}
\[ \frac{B_{a/2}}{B_{2a}} = \frac{\frac{\mu_0 I}{\pi a}}{\frac{\mu_0 I}{4\pi a}} = \frac{1}{1} = 1:1 \] Thus, the correct answer is \( 1:1 \).