The magnetisation at the centre of a bar magnet (magnetic length 10 cm), having pole strength 6.3 A m, and area of cross-section $0.70\, cm^2$ is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $9.0 × 10^4 A m^{-1}$
Given:
Magnetic length, $l = 10\,cm = 0.10\,m$
Pole strength, $m = 6.3\,A·m$
Area of cross-section, $A = 0.70\,cm^{2} = 0.70 \times 10^{-4}\,m^{2} = 7.0 \times 10^{-5}\,m^{2}$
Magnetic moment, $μ = m \times l$
Volume of magnet, $V = A \times l$
Magnetisation, $M = \frac{μ}{V} = \frac{m \times l}{A \times l} = \frac{m}{A}$
Substitute values:
$M = \frac{6.3}{7.0 \times 10^{-5}}$
$M = 9.0 \times 10^{4}\,A/m$
Final Answer: $M = 9.0 \times 10^{4}\,A/m$