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Q.

In an uniform field the magnetic needle completes 10 oscillations in 92 seconds. When a small magnet is placed in the magnetic meridian 10cm due north of needle with north pole towards south completes 15 oscillations in 69 seconds. The magnetic moment of magnet is ………… A-m2 (given BH = 0.3G)

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answer is 0.75.

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Detailed Solution

n=12πMBI 1092 α BH .............(1) 1569 α B−BH .............(2) From (1) & (2) ⇒ B=5BH We know that, B=μo4π2Md3 ⇒5BH=10−72M(10 x 10−2)3 ⇒5 x 0.3 x 10−4=10−7x103x2M ⇒M=0.75Am2
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