1.

A bar magnet of magnetic moment `1.5JT^-1` lies aligned with the direction of a uniform magnetic field of `0.22T`. (a) What is the amount of work done 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)?

Answer» Here, `M=1.5JT^-1`, `B=0.22T`, `W=?`
(a) (i) Here, `theta_1=0^@` (along the field), `theta_2=90^@` (`_|_` to the field)
As `W=-MB(costheta_2-costheta_1)`
`:. W=-1.5xx0.22(cos 90^@-cos0^@)=-0.33(0-1)=0.33J`
(ii) Here, `theta_1=0^@`, `theta_2=180^@`
`W=-1.5xx0.22(cos 180^@-cos0^@)=-0.33(-1-1)=0.66J`
(b) Torque `tau=MB sin theta`
(i) Here, `theta=90^@`, `tau=1.5xx0.22sin90^@=0.33Nm`
(ii) Here, `theta=180^@`, `tau=1.5xx0.22sin180^@=0`


Discussion

No Comment Found

Related InterviewSolutions