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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

`6.5 xx 10^(-5)N`

B

`3.6 xx 10^(-5)N`

C

`1.3 xx 10^(-5)N`

D

`1.8 xx 10^(-5)N`

Text Solution

Verified by Experts

The correct Answer is:
A
At `45^(@) , B_(H) = B_(v)`
`(Fl)/2 =MB_(v) = m xx l xx B_(v)`
`F = (2mlB_(v))/l = 3.6 xx 18 xx 10^(-6) N `
`= 6.5 xx 10^(-5) `N .
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