Question

An iron rod is placed parallel to magnetic field intensity 1000 A/m. The magnetic flux through the rod is \(3\times 10^{-4}\) Wb and its cross-sectional area is \(1.5 \text{cm}^2\). The magnetic permeability of rod in \(\text{Wb/}_{\text{A-m}}\) is

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

Hint:

Flux density B = flux / area, and permeability mu = B / H.

Answer 1

The Correct Option is B

Step 1: Convert area
$1.5 \text{ cm}^2 = 1.5 \times 10^{-4} \text{ m}^2$.

Step 2: Field
$B = 3\times10^{-4}/1.5\times10^{-4} = 2$ T.

Step 3: Permeability
$\mu = 2/1000 = 2\times10^{-3}$. Option (B).

Final Answer:
Option (B). \[ \boxed{2 \times 10^{-3}} \]