Question

A rod with a circular cross-section area \(2\) cm\(^2\) and length \(40\) cm is wound uniformly with \(400\) turns of an insulated wire. If a current \(0.4\) A flows in the wire winding, the total magnetic flux produced inside the winding is \(4π\times 10^{-6}\) Wb. The relative permeability of the rod is
(Given permeability of vacuum \(μ_0 = 4π\times 10^{-7}\) N m A\(^{-2}\))

• A. \(12.5\)
• B. \(32/5\)
• C. \(125\)
• D. \(5/16\)

Hint:

\(B=\mu_0\mu_r nI\) and \(B=\Phi/A\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Inside a solenoid with a core, \(B = \mu_0\mu_rnI\), where \(n\) is turns per metre. Flux \(\Phi = BA\).

Step 2: Find B and n:
\(A = 2\) cm\(^2 = 2\times10^{-4}\) m\(^2\). \(B = \frac{\Phi}{A} = \frac{4\pi\times10^{-6}}{2\times10^{-4}} = 2\pi\times10^{-2}\) T.
\(n = \frac{400}{0.4} = 1000\) turns/m.

Step 3: Solve:
\(\mu_0nI = 4\pi\times10^{-7}\times1000\times0.4 = 1.6\pi\times10^{-4}\) T.
\[ \mu_r = \frac{2\pi\times10^{-2}}{1.6\pi\times10^{-4}} = 125 \]

Final Answer:
The relative permeability of the rod is \(125\), option (C). \[ \boxed{125} \]