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A vibration magnetometer placed in a magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of 2sec in the earth's horizontal magnetic field of 24 microteslas. When a horizontal field of 18 microteslas is produced opposite to the earth's field by placing a current-carrying wire, the new time period of the magnet will be


A

1s

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B

2s

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C

3s

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D

4s

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Solution

The correct option is D

4s


Step 1: Given data

The magnet executes oscillations with a time period of 2sec.

The earth's horizontal magnetic field of 24 microteslas.

A horizontal field of 18 microteslas.

Step 2: To find

We have to determine the new time period of the magnet.

Step 3: Calculation

We know that,

The time period T of oscillation of a magnet is:

T=2πIMB

Here,

l is the moment of inertia of the magnet about the axis of rotation.

M is the magnetic moment of the magnet.

B is the uniform magnetic field as l, remains the same.

Now,

T1B or, T2T1=B1B2

Then,

B1=24utB2=24ut18ut=6utT1=2secondsT2=224ut6ut=24=2×2=4s

Therefore, option D is the correct choice.


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