A rod (length = 10 cm) moves on two horizontal frictionless conducting rails, as

Question

A rod (length = 10 cm) moves on two horizontal frictionless conducting rails, as shown. The magnetic field in the region is directed perpendicularly to the plane of the rails and is uniform and constant. If a constant force of 0.60 N moves the bar at a constant velocity 2.0 m/s. what is the current through the 12-Ohm load resistor? a. 0.32 A b. 0.34 A c. 0.37 A d. 0.39 A e. 0.43 A For the circuit shown below, find the current in the 20-Ohm resistor and the potential difference between points a and b: A. I_20 Ohm = 0.027A, |V_a - V_b| = 0.91 V B. I_20 Ohm = 0.227A, |V_a - V_b| = 5.68 V C. I_20 Ohm = 0.227A, |V_a - V_b| = 22.8 V D. I_20 Ohm = 2.507A, |V_a - V_b| = 50.1 A rectangular coil (0.20 m times 0.80 m) has 200 turns and is located in a uniform magnetic field of 0.30 T. When the orientation of the coil is varied through all possible positions, the maximum torque on the coil by magnetic forces is 0.080 N middot m. The measure of the current in the coil is closest to: A. 42.0 mA B. 1.7 A C. 5..0 mA D. 1.0 A E. 8.3 mA

Explanation / Answer

4) F = ILB

B = F/IL

V = IR = BvL = FVL/IL

12*I = 0.6*2/I

I = 0.32 A

correct option is (a)

5)

R1 = 20 ohms ,R2 = 10 ohms

R3 = 10 ohms ,R4 = 5ohms ,R5 = 10 ohms

R12 = 5+20 = 25 ohms

1/R1234 = 1/25 +1/10+1/5

R1234 = 2.94 ohms

R12345 = 10+2.94 = 12.94 ohms

V =25 V

I12345 = 25/12.94 = 1.93 A

V5 = 1.93*10 = 19.3 V

Vab = 25 -19.3 = 5.7 V

correct option is (b)

6) N = 200 turns , A = 0.2m *0.8 m

B = 0.3 T

T = = 0.08 N.m

T = NABI

0.08 = 200*0.2*0.8*0.3*I

I = 8.3 mA

correct option is (E)

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