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प्रश्न
A short bar magnet is placed in an external magnetic field of 700 gauss. When its axis makes an angle of 30° with the external magnetic field, it experiences a torque of 0.014 Nm. Find the magnetic moment of the magnet, and the work done in moving it from its most stable to the most unstable position.
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उत्तर
Data: B = 700 gauss = 0.07 tesla, θ = 30°, τ = 0.014 Nm, τ = MB sin θ
The magnetic moment of the magnet is
M = `tau/("B sin" theta) = (0.014)/((0.07)(sin 30^circ)) = 0.4 "A.m"^2`
The most stable state of the bar magnet is for θ = 0°. It is in the most unstable state when θ = 180°. Thus, the work done in moving the bar magnet from 0° to 180° is
W = MB (cos θ0 - cos θ)
= MB (cos 0° - cos 180°)
= MB [1 - (- 1)]
= 2 MB
= (2)(0.4)(0.07)
= 0.056 J
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