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 the magnetic field at distances \(\frac{a}{2}\) and \(2a\) from the axis of a long straight wire carrying a steady current \(I\), we use Ampere's Law. This law relates the integrated magnetic field around a closed loop to the electric current passing through the loop.

Ampere's Law is given by:

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

Where \(\mathbf{B}\) is the magnetic field, \(d\mathbf{l}\) is the differential length element of the loop, \(\mu_0\) is the permeability of free space, and \(I_{\text{enclosed}}\) is the current enclosed by the loop.

• Inside the wire (at \(r = \frac{a}{2}\)):
  • Current density, \(J = \frac{I}{\pi a^2}\), since the current is uniformly distributed.
  • Current enclosed, \(I_{\text{enclosed}} = J \times \pi \left(\frac{a}{2}\right)^2 = \frac{I}{\pi a^2} \times \frac{\pi a^2}{4} = \frac{I}{4}\).
  • Applying Ampere's Law: \(\oint \mathbf{B} \cdot d\mathbf{l} = B\left(2\pi \frac{a}{2}\right) = \mu_0 \frac{I}{4}\).
  • Simplifying gives the magnetic field, \(B_{\frac{a}{2}} = \frac{\mu_0 I}{4\pi a}\).
• Outside the wire (at \(r = 2a\)):
  • The entire current \(I\) is enclosed since \(r = 2a\) is outside the wire.
  • Applying Ampere's Law: \(\oint \mathbf{B} \cdot d\mathbf{l} = B(2\pi \times 2a) = \mu_0 I\).
  • Simplifying gives the magnetic field, \(B_{2a} = \frac{\mu_0 I}{4\pi a}\).

Now, calculate the ratio of magnetic fields:

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

The ratio of the magnetic field at \(\frac{a}{2}\) to that at \(2a\) is therefore 1 : 1.

Answer 2

For a point inside the wire (\(r<a\)), the magnetic field is given by:
\[B_1 \times 2\pi \frac{a}{2} = \mu_0 \frac{I}{4} \implies B_1 = \frac{\mu_0 I}{4\pi a}.\]
For a point outside the wire (\(r>a\)), the magnetic field is:
\[B_2 \times 2\pi \times 2a = \mu_0 I \implies B_2 = \frac{\mu_0 I}{4\pi a}.\]
The ratio of magnetic fields:
\[\frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{4\pi a}}{\frac{\mu_0 I}{4\pi a}} = 1 : 1.\]