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A magnetic needle is placed in a uniform magnetic field and is aligned with the field. The needle is now rotated by an angle of 60^(@) and the work done is W. The torque on the magnetic needle at this position is |
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Answer» `2sqrt3W` where M is the magnetic moment of the magnetic needle. Here, `theta_(1) = 0^(@),theta_(2) = theta = 60^(@)` `therefore W = MB(cos0^(@) - cos60^(@))` `W = MB (1-1/2)=(MB)/2` .....(i) Torque on the needle is `vectau=vecM XX VECB` In magnitude, `tau = MBsintheta` `tau = MB SIN 60^(@) = sqrt3/2MB` .....(ii) Dividing (ii) by (i), we get `tau/W = sqrt3/2 MB xx 2/(MB) = sqrt3 or tau = sqrt3W` |
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