Question

An iron rod is placed parallel to magnetic field of intensity H. The magnetic flux through the rod is \(\Phi\) and its cross-sectional area is A. The magnetic permeability (\(\mu\)) of the rod is

• A. \(\Phi AH\)
• B. \(\frac{AH}{\Phi }\)
• C. \(\frac{\Phi }{AH}\)
• D. \(A/\Phi H\)

Hint:

Flux is B times area, and B equals permeability times H.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Magnetic flux through the rod is \(\Phi=BA\) since the field is along the rod.

Step 2: Relate B and H
Inside the material \(B=\mu H\).

Step 3: Combine
\(\Phi=\mu HA\).

Step 4: Solve
\(\mu=\frac{\Phi}{AH}\), option (C).

Final Answer:
The permeability is flux divided by the product of area and H. \[ \boxed{\text{(C) }\dfrac{\Phi}{AH}} \]