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19/09/2020 5:55 pm
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An iron rod of volume 10–3m3 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 :
(1) 0.5 × 102 Am2
(2) 5 × 102 Am2
(3) 500 × 102 Am2
(4) 50 × 102 Am2
1 Answer
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19/09/2020 5:59 pm
(2) 5 × 102 Am2
Explanation:
Magnetic moment of an iron core solenoid
M = (μr - 1).NiA
= (μr - 1).Ni v/ℓ
= (μr - 1).Ni/ℓ iV
= 999 x 10/ 10–2 x 0.5 x 10–3
= 499.5 ≈ 500
This post was modified 4 years ago by nikhil04