Question

A short bar magnet is placed with its south pole towards geographical north. Neutral points are at 20 cm from centre. Given \(B_H = 0.3\times10^{-4}\,Wb/m^2\), find magnetic moment.

• A. \(9000\,\text{ab-amp}\times\text{cm}^2\)
• B. \(900\,\text{ab-amp}\times\text{cm}^2\)
• C. \(1200\,\text{ab-amp}\times\text{cm}^2\)
• D. \(225\,\text{ab-amp}\times\text{cm}^2\)

Hint:

Neutral point $\Rightarrow$ field of magnet equals horizontal component of Earth’s field.

Answer 1

The Correct Option is C

Concept: At neutral point: \[ B_{\text{magnet}} = B_H \] For axial line: \[ B = \frac{\mu_0}{4\pi}\frac{2M}{r^3} \]

Step 1: Equate fields
\[ \frac{\mu_0}{4\pi}\frac{2M}{r^3} = B_H \]

Step 2: Substitute values
\[ \frac{10^{-7}\cdot 2M}{(0.2)^3} = 0.3\times10^{-4} \] \[ \frac{2\times10^{-7}M}{8\times10^{-3}} = 0.3\times10^{-4} \] \[ \frac{2M}{8}\times10^{-4} = 0.3\times10^{-4} \] \[ \frac{M}{4} = 0.3 \Rightarrow M = 1.2\,A\,m^2 \]

Step 3: Convert units
\[ 1\,A\,m^2 = 10^3\,\text{ab-amp}\times\text{cm}^2 \] \[ M = 1.2\times10^3 = 1200\,\text{ab-amp}\times\text{cm}^2 \] Conclusion: \[ 1200\,\text{ab-amp}\times\text{cm}^2 \]