Question

when the intensity of magnetic field is increased four times the time period of suspended magnetic needle becomes

Answer 1

Hey, there!

Here the answer.

$T = \frac{2m\pi}{qB}$
So if B is increased 4times, then

${T}^{'} = \frac{2m\pi}{q(4B)} = \frac{T}{4}$
Therefore, the time period decreases 4times.


Hope that helps.

Thanks.

Answer 2

time period of suspended needle becomes one fourth of its initial value.

when a charge q of mass m is subjected in the magnetic field B with velocity v.

then, centripetal force = magnetic force

⇒mv²/r = Bqv

⇒mv/r = Bq

⇒mω = Bq

⇒m(2π/T) = Bq

⇒T = 2πm/qB

here it is clear that time period of charged particle is inversely proportional to intensity of magnetic field.

so, if intensity of magnetic field is increased four time, time period of suspended magnetic needle becomes one fourth of its initial value.

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