A conducting rod of length 50.0 cm slides over two parallel metal bars placed in

Question

A conducting rod of length 50.0 cm slides over two parallel metal bars placed in a magnetic field with a magnitude of 1000. G, as shown in the figure. The ends of the rods are connected by two resistors, R = 100. Q and R2 - 200. Q. The conducting rod moves with a constant speed of 8.00 m/s. What are the currents flowing through the two resistors? A rectangular coil with 20 windings carries a current of 2.00 mA flowing in the counterclockwise direction. It has two sides that are parallel to they-axis and have length 8.00 cm and two sides that are parallel to the *-axis and have length 6.00 cm. A uniform magnetic field of 50.0 fiT acts in the positive jc-direction. What torque must be applied to the loop to hold it steady?

Explanation / Answer

6.

B = magnetic field = 1000 gauss = 1000 x 10-4 T = 0.1 T                  since 1 G = 10-4 T

L = length of rod = 50 cm = 0.50 m

V = speed of rod = 8 m/s

induced EMF in the rod is given as

E = BL V = 0.1 x 0.50 x 8 = 0.4 volts

Net resistance = R1 R2 / (R1 + R2)                     since R1 and R2 are in parallel

R = 100 x 200 / (100 + 200)

R = 66.67 ohm

current coming from the Voltage = i = E/R = 0.4 / 66.67 = 6 x 10-3 A

current in R1 = i1 = E/R1 = 0.4 / 100 = 0.004 A

current in R2 = i2 = E/R1 = 0.4 / 200 = 0.002 A

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