Question

An iron rod is placed parallel to magnetic field intensity 2000 A/m. The magnetic flux through rod is \(6\times 10^{-4}\) wb and its cross sectional area is 3 cm\(^2\). The magnetic permeability of the rod in wb/A-m is

• A. \(10^{-1}\)
• B. \(10^{-2}\)
• C. \(10^{-3}\)
• D. \(10^{-4}\)

Hint:

Find the flux density B from flux over area, then use mu = B/H.

Answer 1

The Correct Option is C

Step 1: Combine the formulas.
$\mu = \dfrac{\Phi}{A H}$.

Step 2: Substitute.
$\mu = \dfrac{6\times10^{-4}}{(3\times10^{-4})(2000)} = \dfrac{6\times10^{-4}}{0.6}$.

Step 3: Simplify.
$= 10^{-3}$ Wb/A m.

Final Answer:
Option (C). \[ \boxed{10^{-3}\text{ Wb/A m}} \]