JEE Mains · Physics · STD 12 - 5. Magnetism and matter
An iron rod of volume \(10^{-3}\, m ^{3}\) 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\(...........Am^2\)
- A \(0.5 \times 10^{2}\)
- B \(50 \times 10^{2}\)
- C \(500 \times 10^{2}\)
- D \(5 \times 10^{2}\)
Answer & Solution
Correct Answer
(D) \(5 \times 10^{2}\)
Step-by-step Solution
Detailed explanation
\(M =\mu_{ r } Ni A\) Here \(\mu_{ r }=\) Relative permeability \(N =\) No. of turns \(i =\) Current \(A =\) Aea of cross section \(M=\mu_{r} N i A=\mu_{r} n\ell i A\) \(M=\mu_{r} n i V=1000(1000) 0.5\left(10^{-3}\right)\) \(=500=5 \times 10^{2} Am ^{2}\)
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