Question

A bar magnet having length $5\,\text{cm$ and area of cross-section $4\,\text{cm}^2$ has magnetic moment $2\,\text{Am}^2$. If magnetic susceptibility is $5 \times 10^{-6}$, the magnetic intensity will be}

• A. $0.2 \times 10^{10} \frac{A}{m}$
• B. $0.5 \times 10^{10} \frac{A}{m}$
• C. $5 \times 10^{10} \frac{A}{m}$
• D. $2 \times 10^{10} \frac{A}{m}$

Hint:

Magnetic intensity depends on the ratio of magnetic moment and volume of the magnet.

Answer 1

The Correct Option is D

Step 1: Magnetic intensity relation.
Magnetic intensity $H$ is related to magnetic moment $M$ and volume $V$ of the bar magnet by: \[ H = \frac{M}{V} \]
Step 2: Calculating volume of the magnet.
The volume of the bar magnet is: \[ V = \text{Length} \times \text{Area} = 5 \times 4 = 20\,\text{cm}^3 = 20 \times 10^{-6}\,\text{m}^3 \]
Step 3: Substituting values.
Substitute $M = 2\,\text{Am}^2$ and $V = 20 \times 10^{-6}\,\text{m}^3$: \[ H = \frac{2}{20 \times 10^{-6}} = 2 \times 10^{10} \frac{A}{m} \]
Step 4: Conclusion.
The magnetic intensity is $2 \times 10^{10} \frac{A}{m}$.