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In a vibration magnetometer, the time pe...

In a vibration magnetometer, the time period of a bar magnet oscillating in horizontal componnt of earth's magnetic field is `2 sec`. When a magnet is brought near and parallel to it, the time period reduces to `1 sec`. The ratio `H//F` of the horizontal component `H` and the field `F` due to magnet will be

A

3

B

`1//3`

C

`sqrt3`

D

`1//sqrt3`

Text Solution

Verified by Experts

The correct Answer is:
B
`T prop (1)/(sqrtH) rArr (T_(1))/(T_(2)) = sqrt((H_(2))/(H_(1)))`
`((2)/(1))^(2)= (H + F)/(H) rArr F= 3H,(H)/(F)= (1)/(3)`
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