A bar magnet with poles ` 25 cm` apart and of strength `14.4 amp - m` rests with centre on a frictionless pivot. It is held in equilibrium aat an angle of ` 60@` with respect to a uniform magnetic field of induction `0.25 Wb//m^(2) , by applying a force `F` at right angles to its axis at a point `12 cm` from pivot. Calculate `F`. What will happen if the force `F` is removed?
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The correct Answer is:
B
` 2l = 0.25 m` Also, ` mxx2l = 14.4 rArr m = (14.4)/(0.25) = 57.6A-m^(2)` Torque due to magnetic field ` = p_(m) xxB xxsin60^(@) = 14.4xx0.25xx (sqrt(3))/(2)` The torque due to the force ` = Fxx0.12` For equilibrium `Fxx0.12 = 14.4xx0.25xx(sqrt(3))/(2) rArr F = 25.98N` If the force `F` is removed , the torque due to magnetic field will move the bar magnet . It will start oscillating about the mean position where the angle between ` vec(pP_(m) and vec(B)` is `0`.
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