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[物理]普通物理的問題!!! (電磁感應)

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發問:

A square coil has turns of side 5 cm and a resistance of 2.5 Ω.It rotates at 120rpm about a vertical axis in a horizontal magnetic field of0.04T. At t = 0, the plane of the coil is perpendicular to the field,At t = 0.1 s find;(a) the induced emf;(b) the torque needed to keep the coil... 顯示更多 A square coil has turns of side 5 cm and a resistance of 2.5 Ω. It rotates at 120rpm about a vertical axis in a horizontal magnetic field of 0.04T. At t = 0, the plane of the coil is perpendicular to the field, At t = 0.1 s find; (a) the induced emf; (b) the torque needed to keep the coil rotating; (c) the machanical power needed to keep the coil rotating; (d) the electrical power dissipated. 可以幫我這個普物的問題嗎? 謝謝

最佳解答:

Suppose that the coin has N turns with its orientation making an angle of θ with the vertical, then its flux linkage is: NΦ = NBA = N (0.04 0.05 0.05) cos θ = 10-4 N cos θ (Wb) Now, with dθ/dt = 120 2π/60 = 4π rad/sec, we have: NΦ = 10-4 N cos 4πt (a) Induced emf E = -d(NΦ)/dt = -4π 10-4 N sin 4πt with t = 0.1, E = 1.195 10-3 N (V) (b) At t = 0.1, current in the coil = 1.195 10-3 N/2.5 = 4.781 10-4 N (A) So the torque due to this current is NBAIcos4πt = 1.477 10-8 N2 (Nm) which is also the value needed for keeping the coil rotating. (c) By power = Torque Angular speed, we have: Power = 1.477 10-8 N2 4π = 1.856 10-7 N2 (W) (d) By P = V2/R, we have: P = 5.712 10-7 N2 (W)

其他解答:BFC66BE0445C3814

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