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A bar magnet of magnetic moment 1.5 JT^(-1) lies aligned with the direction of a uniform magnetic field of 0.22 T. (a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction (ii) opposite to the field direction? (b) What is the torque on the magnet in cases (i) and (ii)? |
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Answer» Solution :(a) (i) Required amount of work done, `W= mB ( cos theta_(1)- cos theta_(2) )` `= (1.5 ) (0.22) ( cos 0^(@) -cos 90^(@) )` `=(1.5)(0.22) (1-0)` `=0.33` J (ii) In this case, `W= mB ( cos theta_(1)- cos theta_(2))` `THEREFORE W= (1.5) ( 0.22)( cos0^(@) - cos180^(@) )` `therefore= (1.5) (0.22) (1- ( -1) )` `therefore W= (1.5) (0.22) (2)` `therefore W= 0.66` J (B) (i) Torque exerted, `tau = m B sin theta` `therefore tau = (1.5) (0.22) sin 90^@` `therefore tau= (1.5) (0.22) (1)` `therefore tau = 0.33` Nm (ii) In this case, `tau= m B sin theta` `therefore tau= (1.5) (0.22) (0)` `therefore tau=0` |
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