Question

A long straight wire of radius a carries a steady current I. The current uniformly distributed over its cross-section. The ratio of the magnetic fields B and B', at radial distance $\frac{a}{2}$ and 2a respectively, from the axis of the wire is :

• A. $\frac{1}{2} $
• B. 1
• C. 4
• D. $\frac{1}{4}$

Answer 1

The Correct Option is B

Magnetic Field Inside and Outside a Wire 

For a current-carrying wire, the magnetic field at any point depends on the position relative to the wire. The equations for the magnetic field are different for points inside and outside the wire:

1. For Points Inside the Wire

The magnetic field at a point inside the wire (at a distance \( r \) from the center of the wire, where \( r \leq R \), and \( R \) is the radius of the wire) is given by:

\(B = \frac{\mu_0 I r}{2 \pi R^2} \quad (r \leq R)\)

2. For Points Outside the Wire

The magnetic field at a point outside the wire (at a distance \( r \) from the center of the wire, where \( r \geq R \)) is given by:

\(B = \frac{\mu_0 I}{2 \pi r} \quad (r \geq R)\)

3. Ratio of Magnetic Fields

According to the question, we are comparing the magnetic fields at two points:

\(\frac{B}{B'} = \frac{\frac{\mu_0 I \left(\frac{a}{2}\right)}{2 \pi a^2}}{\frac{\mu_0 I}{2 \pi (2a)}}\)

Simplifying the expression:

\(\frac{B}{B'} = \frac{\frac{\mu_0 I (a / 2)}{2 \pi a^2}}{\frac{\mu_0 I}{2 \pi (2a)}} = 1 : 1\)

Conclusion:

The ratio of the magnetic fields at the two points is \( 1 : 1 \), meaning the magnetic fields are equal.