At a point on the surface of earth, time period of oscillation of magnetic needle is T. Now a uniform magnetic field B, perpendicular to earth's horizontal component of magnetic field is produced in that region and time period of oscillation of the same needle at the same place is found to be T3. Then BBH is equal to

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answer is 3.

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Detailed Solution

T=2πIMBH⇒BH=4π2IMT2T1=2πIMB1⇒B1=4π2IMT12=BHT2T12Now, B2+BH2=B1 Given T1=T3∴B2+BH2=BHTT12=BH32=3BH ⇒BBH=22

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