Question:
An iron rod of volume \(10^{-3}\) m3 and relative permeability 1000 is placed as core in a solenoid with 10 turns/cm. If a current of 0.5 A is passed through the solenoid, then the magnetic moment of the rod will be :
An iron rod of volume \(10^{-3}\) m3 and relative permeability 1000 is placed as core in a solenoid with 10 turns/cm. If a current of 0.5 A is passed through the solenoid, then the magnetic moment of the rod will be :
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Magnetic moment m = M V, with \(M = \chi H = (\mu_r - 1)nI\) and n = 1000 turns/m.
Updated On: Nov 5, 2026
- \(5\times 10^{2}\) Am2
- \(0.5\times 10^{2}\) Am2
- \(500\times 10^{2}\) Am2
- \(50\times 10^{2}\) Am2
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The Correct Option is A
Solution and Explanation
Turns per metre: n = 10 turns/cm = 1000 turns/m.
Magnetising field: H = nI = 1000 \(\times\) 0.5 = 500 A/m.
Susceptibility: \(\chi = \mu_r - 1 = 999 \approx 1000\).
Magnetisation: M = \(\chi H\) = 999 \(\times\) 500 \(\approx 5\times 10^5\) A/m.
Magnetic moment: m = M V = \(5\times 10^5 \times 10^{-3}\) = 500 Am2 = \(5\times 10^2\) Am2.
Answer: (1) \(5\times 10^{2}\) Am2
Magnetising field: H = nI = 1000 \(\times\) 0.5 = 500 A/m.
Susceptibility: \(\chi = \mu_r - 1 = 999 \approx 1000\).
Magnetisation: M = \(\chi H\) = 999 \(\times\) 500 \(\approx 5\times 10^5\) A/m.
Magnetic moment: m = M V = \(5\times 10^5 \times 10^{-3}\) = 500 Am2 = \(5\times 10^2\) Am2.
Answer: (1) \(5\times 10^{2}\) Am2
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