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18/09/2020 4:20 pm
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A small bar magnet is placed with its axis at 30° with an external magnetic field of 0.06 T experiences a torque of 0.018 Nm. The minimum work required to rotate it from its stable to unstable equilibrium position is:
(1) 9.2 × 10-3 J
(2) 11.7 × 10-3 J
(3) 6.4 × 10-2 J
(4) 7.2 × 10-2 J
1 Answer
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18/09/2020 4:26 pm
(4) 7.2 × 10-2 J
Explanation:
τ = MBsinθ = 0.18
ω = ΔU = Ur -Ui
= –MBcos180° – (–MBcos0°)
= 2MB
= 2 × 0.6 × 0.06
= 0.072 J