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: Understanding the Concept:
Flux density is \(B = \dfrac{\Phi}{A}\) and the permeability of a material is \(\mu = \dfrac{B}{H}\).

Step 2: Convert the units.
\(A = 3\text{ cm}^2 = 3\times 10^{-4}\text{ m}^2\).

Step 3: Find B.
\[ B = \frac{6\times 10^{-4}}{3\times 10^{-4}} = 2\text{ Wb/m}^2 \]

Step 4: Find \(\mu\).
\[ \mu = \frac{B}{H} = \frac{2}{2000} = 10^{-3}\text{ Wb/A m} \]

Final Answer:
The permeability is \(10^{-3}\) Wb/A-m, option (C). \[ \boxed{10^{-3}\text{ Wb/A m}} \]