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A bar magnet with poles 25.0 apart and of polestrength 14.4 am rests with its cent4re on a frictionless point12 cm from its pivot the magnitude of the force is |
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Answer» `15sqrt(3) N` `Fxx0.12 =MB sin THETA` `F=(3.6xx0.25 xxsqrt(3)//2)/(0.12)` `=3.75 sqrt(3)N`
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