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: Understanding the Concept
Magnetic flux density is \(B = \dfrac{\phi}{A}\), and permeability is \(\mu = \dfrac{B}{H}\).

Step 2: Compute
Area \(A = 1.5\text{ cm}^2 = 1.5 \times 10^{-4}\text{ m}^2\).
\[ B = \frac{3 \times 10^{-4}}{1.5 \times 10^{-4}} = 2 \text{ T} \]
\[ \mu = \frac{B}{H} = \frac{2}{1000} = 2 \times 10^{-3} \text{ Wb/(A m)} \]
Option (A) results if the area is not converted to square metres correctly.

Final Answer:
The permeability is \(2 \times 10^{-3}\) Wb/A-m, option (B). \[ \boxed{2 \times 10^{-3}} \]