A metal rod is forced to move with constant velocity along two parallel metal rails, connected with a strip of metal at one end across a magnetic field (B) of 0.5 T, pointing out of the page. The rod is of length 45 cm and the speed of the rod is 70 cm/s. The rod has a resistance of 10 Ω and the resistance of the rails and connector is negligible. What is the rate at which energy is being transferred to thermal energy?(a) 0.225 W(b) 22.55 W(c) 2.25 × 10^-4 W(d) 2.25 × 10^-3 WThis question was addressed to me in homework.Origin of the question is Electromagnetic Induction in section Electromagnetic Induction of Physics – Class 12

A metal rod is forced to move with constant velocity along two paralle

1.	A metal rod is forced to move with constant velocity along two parallel metal rails, connected with a strip of metal at one end across a magnetic field (B) of 0.5 T, pointing out of the page. The rod is of length 45 cm and the speed of the rod is 70 cm/s. The rod has a resistance of 10 Ω and the resistance of the rails and connector is negligible. What is the rate at which energy is being transferred to thermal energy?(a) 0.225 W(b) 22.55 W(c) 2.25 × 10^-4 W(d) 2.25 × 10^-3 WThis question was addressed to me in homework.Origin of the question is Electromagnetic Induction in section Electromagnetic Induction of Physics – Class 12
Answer» Right choice is (d) 2.25 × 10^-3 W Easy explanation: Given: B = 0.5 T; v = 70 cm/s = 70 × 10^-2 m/s; L = 45 cm = 45 × 10^-2 m; R = 10 Ω Motional EMF (E) = vLB = 70 × 10^-2 × 45 × 10^-2 × 0.5 E = 0.15 V Current (I) = \(\frac {E}{R}\) I = \(\frac {0.15}{10}\) I = 0.015 A RATE of ENERGY or power (P) = I^2R P = 0.015^2 × 10 P = 2.25 × 10^-3 W Therefore, the rate of energy TRANSFER is 2.25 × 10^-3 W.

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