標題:
Emf problem
發問:
A metal rod is pulled by a force F at a constant velocity of 50 m/s along the frictionless limbs of a U-shaped metal wire as shown in the figure. The magnetic field has a flux density of 0.35 T and points out of the page. The limbs are separated by 24 cm and the total resistance is 10 Ω. Find (i) the... 顯示更多 A metal rod is pulled by a force F at a constant velocity of 50 m/s along the frictionless limbs of a U-shaped metal wire as shown in the figure. The magnetic field has a flux density of 0.35 T and points out of the page. The limbs are separated by 24 cm and the total resistance is 10 Ω. Find (i) the induced e.m.f. across the rod, (ii) the induced current in the rod, (iii) the power dissipated in the 10 ohm resistor, (iv) the force F required to keep the rod moving at constant velocity, (v) and the rate at which internal energy is generated. http://i.imgur.com/2JWtnpv.png
最佳解答:
(i) Induced emf = 0.35 x 0.24 x 50 v = 4.2 v (ii) Induced current = 4.2/10 A = 0.42 A (iii) Power dissipated = 0.42^2 x 10 w = 1.764 w (iv) Force required = 0.35 x 0.24 x 0.42 N = 0.03528 N (v) Power = 0.03528 x 50 w = 1.764 w
Emf problem
發問:
A metal rod is pulled by a force F at a constant velocity of 50 m/s along the frictionless limbs of a U-shaped metal wire as shown in the figure. The magnetic field has a flux density of 0.35 T and points out of the page. The limbs are separated by 24 cm and the total resistance is 10 Ω. Find (i) the... 顯示更多 A metal rod is pulled by a force F at a constant velocity of 50 m/s along the frictionless limbs of a U-shaped metal wire as shown in the figure. The magnetic field has a flux density of 0.35 T and points out of the page. The limbs are separated by 24 cm and the total resistance is 10 Ω. Find (i) the induced e.m.f. across the rod, (ii) the induced current in the rod, (iii) the power dissipated in the 10 ohm resistor, (iv) the force F required to keep the rod moving at constant velocity, (v) and the rate at which internal energy is generated. http://i.imgur.com/2JWtnpv.png
最佳解答:
(i) Induced emf = 0.35 x 0.24 x 50 v = 4.2 v (ii) Induced current = 4.2/10 A = 0.42 A (iii) Power dissipated = 0.42^2 x 10 w = 1.764 w (iv) Force required = 0.35 x 0.24 x 0.42 N = 0.03528 N (v) Power = 0.03528 x 50 w = 1.764 w
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