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 \(\frac{a}{2}\) and \(2a\) from axis of the wire is:

• A. 1 : 4
• B. 4 : 1
• C. 1 : 1
• D. 3 : 4

Answer 1

The Correct Option is C

To determine the ratio of magnetic field strength at radial distances of \(\frac{a}{2}\) and \(2a\) from the axis of a long straight wire carrying a steady current \(I\), Ampere's Law is applied. Ampere's Law establishes a relationship between the integral of the magnetic field along a closed loop and the net electric current traversing that loop.

The mathematical formulation of Ampere's Law is:

\(\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{enclosed}}\)

In this equation, \(\mathbf{B}\) represents the magnetic field, \(d\mathbf{l}\) is an infinitesimal element of the loop's path, \(\mu_0\) denotes the magnetic permeability of free space, and \(I_{\text{enclosed}}\) is the total current encircled by the chosen loop.

• Within the wire (at \(r = \frac{a}{2}\)):
  • The current density, \(J\), is uniform and calculated as \(J = \frac{I}{\pi a^2}\).
  • The enclosed current, \(I_{\text{enclosed}}\), is the current density multiplied by the area of the loop at \(r = \frac{a}{2}\): \(I_{\text{enclosed}} = J \times \pi \left(\frac{a}{2}\right)^2 = \frac{I}{\pi a^2} \times \frac{\pi a^2}{4} = \frac{I}{4}\).
  • Applying Ampere's Law with a circular loop of radius \(\frac{a}{2}\): \(\oint \mathbf{B} \cdot d\mathbf{l} = B\left(2\pi \frac{a}{2}\right) = \mu_0 \frac{I}{4}\).
  • Solving for the magnetic field at this radius yields: \(B_{\frac{a}{2}} = \frac{\mu_0 I}{4\pi a}\).
• Outside the wire (at \(r = 2a\)):
  • Since \(r = 2a\) is beyond the wire's radius, the entire current \(I\) is enclosed by the loop.
  • Applying Ampere's Law with a circular loop of radius \(2a\): \(\oint \mathbf{B} \cdot d\mathbf{l} = B(2\pi \times 2a) = \mu_0 I\).
  • Solving for the magnetic field at this radius gives: \(B_{2a} = \frac{\mu_0 I}{4\pi a}\).

The ratio of the magnetic fields is then computed as:

\(\frac{B_{\frac{a}{2}}}{B_{2a}} = \frac{\frac{\mu_0 I}{4\pi a}}{\frac{\mu_0 I}{4\pi a}} = 1 : 1\)

Consequently, the magnetic field at a distance \(\frac{a}{2}\) from the wire's axis is in a 1 : 1 ratio with the magnetic field at a distance \(2a\).