A \( 30\ \mathrm{cm} \) long solenoid having \( 500 \) turns and radius \( 3\ \mathrm{cm} \) carries a current of \( 15\ \mathrm{A} \). Its magnetisation \( |\vec{M}| \) would be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 25000\ \mathrm{A/m} \)
Magnetisation is the net magnetic moment per unit volume. A current-carrying solenoid behaves like a bar magnet whose magnetic moment is \( m = NIA \), and it fills a volume \( V = Al \).
\( \displaystyle M = \frac{m}{V} = \frac{NIA}{Al} = \frac{N}{l}\,I = nI \)
The cross-section \( A \) cancels, so the radius (3 cm) is not needed.
\( \displaystyle n = \frac{N}{l} = \frac{500}{0.30\,\mathrm{m}} = \frac{5000}{3}\,\mathrm{m^{-1}} \)
\( \displaystyle M = \frac{5000}{3}\,\mathrm{m^{-1}}\times15\,\mathrm{A} = 25000\,\mathrm{A/m} \)
Note: The question does not say what fills the solenoid. Magnetisation normally describes a material placed inside the solenoid. With an air core it is practically zero, because the susceptibility of air is about \( 10^{-7} \), and \( nI = 25000\,\mathrm{A/m} \) is then the magnetic intensity \( H \) inside the solenoid. The answer follows from treating the solenoid itself as an equivalent magnet and taking its magnetic moment per unit volume.