Question

A steady current \(I\) flows through a long straight wire of radius \(a\). The current is uniformly distributed across its cross-section. The ratio of the magnetic fields due to the wire at distances \(\frac{a}{4}\) and \(3a\) respectively from the axis of the wire is:

• A. \(3:4\)
• B. \(4:3\)
• C. \(2:3\)
• D. \(1:4\)

Hint:

Inside a uniformly current-carrying wire, \(B\propto r\). Outside the wire, \(B\propto \frac{1}{r}\).

Answer 1

The Correct Option is A

Concept:
For a long straight wire with uniformly distributed current: Inside the wire \((r<a)\): \[ B=\frac{\mu_0Ir}{2\pi a^2} \] Outside the wire \((r>a)\): \[ B=\frac{\mu_0I}{2\pi r} \]

Step 1: Find magnetic field at \(r=\frac{a}{4}\).
Since \(\frac{a}{4}<a\), the point is inside the wire. \[ B_1=\frac{\mu_0I r}{2\pi a^2} \] \[ B_1=\frac{\mu_0I\left(\frac{a}{4}\right)}{2\pi a^2} \] \[ B_1=\frac{\mu_0I}{8\pi a} \]

Step 2: Find magnetic field at \(r=3a\).
Since \(3a>a\), the point is outside the wire. \[ B_2=\frac{\mu_0I}{2\pi(3a)} \] \[ B_2=\frac{\mu_0I}{6\pi a} \]

Step 3: Find the ratio.
\[ \frac{B_1}{B_2}=\frac{\frac{\mu_0I}{8\pi a}}{\frac{\mu_0I}{6\pi a}} \] \[ =\frac{1}{8}\times 6 \] \[ =\frac{6}{8} \] \[ =\frac{3}{4} \] Therefore, \[ B_1:B_2=3:4 \]