Question

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

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

Hint:

In magnetic field-related questions, always use the correct formula relating flux, field, and permeability.

Answer 1

The Correct Option is C

Step 1: Formula for magnetic permeability.
Magnetic permeability (\( \mu \)) is related to magnetic flux (\( \Phi \)), magnetic field intensity (\( B \)), and cross-sectional area (\( A \)) as: \[ \mu = \frac{\Phi}{B \cdot A} \]
Step 2: Calculation.
The magnetic field intensity \( B \) is given by: \[ B = \frac{2000}{\mu} \] Substitute values into the formula: \[ \mu = \frac{6 \times 10^{-4}}{(2000 \cdot 3 \times 10^{-4})} = 10^{-3} \]
Step 3: Conclusion.
The correct answer is (C) \( 10^{-3} \) as the magnetic permeability of the rod.