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Question : 13
Total: 26
A bar magnet of magnetic moment 6 J ∕ T 60 ∘ 0.44 T
Solution:
(a) Formula and Calculation of work done in the two cases
(b) Calculation of torque in case (ii)
(a) Work done= m B ( cos θ 1 − cos θ 2 )
(i)θ 1 = 60 ∘ , θ 2 = 90 ∘
∴ work done = m B ( cos 60 ∘ − cos 90 ∘ )
= m B (
− 0 ) =
m B
=
× 6 × 0.44 J = 1.32 J
(ii)θ 1 = 60 ∘ , θ 2 = 180 ∘
∴ work done = m B ( cos 60 ∘ − cos 180 ∘ )
= m B (
− ( − 1 ) ) =
m B
=
× 6 × 0.44 J − 3.96 J end a l i g n e d
[Also accept calculations done through changes in potential energy.]
(b) Torque= |
×
| = m B s i n θ
Forθ = 180 ∘
Torque= 6 × 0.44 s i n 180 ∘ = 0
[If the student straight away writes that the torque is zero since magnetic moment and magnetic field are anti parallel in this orientation, award full]
(b) Calculation of torque in case (ii)
(a) Work done
(i)
(ii)
[Also accept calculations done through changes in potential energy.]
(b) Torque
For
Torque
[If the student straight away writes that the torque is zero since magnetic moment and magnetic field are anti parallel in this orientation, award full]
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