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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

Text Solution

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using `Mbsin theta =fl(sin theta (tau)`

`MB sin 45^(@)=F l/2 sin 45^(@)`
`F=2MB=2xx1.8xx10^(-6)=6.5xx10(-5)N`
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