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: {Application of Ampere's Circuital Law}
The magnetic field generated by a straight wire is determined by the formula:\[B = \frac{\mu_0 I}{2\pi r}\]Step 2: {Determination of Magnetic Fields at \( a/2 \) and \( 2a \)}
Magnetic field at distance \( r = a/2 \):\[B_{a/2} = \frac{\mu_0 I}{2\pi (a/2)}\]\[B_{a/2} = \frac{\mu_0 I}{\pi a}\]Magnetic field at distance \( r = 2a \):\[B_{2a} = \frac{\mu_0 I}{2\pi (2a)}\]\[B_{2a} = \frac{\mu_0 I}{4\pi a}\]Step 3: {Calculation of the Ratio}
\[\frac{B_{a/2}}{B_{2a}} = \frac{\frac{\mu_0 I}{\pi a}}{\frac{\mu_0 I}{4\pi a}} = \frac{1}{1} = 1:1\]Therefore, the correct ratio is \( 1:1 \).