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At some location on earth the horizontal...

At some location on earth the horizontal component of earth's magnetic field is `18 xx 10^(-6)T`. At this location, magnetic needle of length 0.12m and pole strength 1.8 Am is supended from its mid-point using a thread, it makes `45^(@)` angle with horizontal in equilibrium. To keep this needle horizontal, th evertical force that should be applied at one of its ends is

A

`6.5xx10^(-5)N`

B

`1.3xx10^(-5)N`

C

`1.8xx10^(-5)N`

D

`3.6xx10^(-5)N`

Text Solution

Verified by Experts

The correct Answer is:
A
As angle of dip `=45^(@)`
`rArr B_(V)=B_(H)=1.8xx10^(-6)`
`tau_(F)=tau_(B)`
`Fxx(l)/(2)=Mxx1.8xx10^(-6) rArr F=3.6xx1.8xx10^(-6)= 6.5xx10^(-5)N`
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