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A short bar magner placed with its axis ...

A short bar magner placed with its axis at `30^(@)` with a uniform external magnetic field of 0.25 T experiences a torque of `4.5 xx 10^(-2)J`. What is its magnetic moment?

Text Solution

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`:.` Torque, `tau=m xx B rArr |tau| =|m| |B| sin theta`
`rArr |B|=` magnetic moment of magnet
`=(|tau|)/( |B| sin theta) =(4.5 xx 10^(-2))/(0.25 xx sin 30^(@))`
`=0.36 "A-m"^(2)`
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