e) 0.250 6. The primary coil in an ideal transformer has N turns. To output half
ID: 1788011 • Letter: E
Question
e) 0.250 6. The primary coil in an ideal transformer has N turns. To output half the input potential difference, the secondary coil should have: a) N/2 turns. b) N turns c) 2N turns. d) a number impossible to determine with the given information 7. Suppose the magnetic field in the figure below has a magnitude of 0.60T, and the length of the conductive rod is 80m. The two conductive rails, rod and wires have no resistance. The resistance R-12. The rod is moving to the right at a constant speed of 3.0m/s (1) What is the motional emf in the moving rod? (2) What is the current in the circuit? (3) what is the direction of the current (CCW or CW)? (4) What is the power delivered to the resistor? (5) What is the magnitude and direction of the magnetic force acting on the moving rod? (6) In order to move the rod at constant speed, what is the magnitude and direction of the applied force?Explanation / Answer
6. (A) N/2 turns
7. (1) emf = B L v
= 0.60 x 0.80 x 3
= 1.44 Volt
(2) I = e / R = 0.12 A
(3) CCW
(4) P = I^2 R = 0.17 W
(5) to the left
F = I L B = 0.12 x 0.80 x 0.60 = 0.0576 N
(6)Fapp = 0.0576 N
to the right
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