A conducting rod of length I = 35.0 cm is free to slide on two parallel conducti

Question

A conducting rod of length I = 35.0 cm is free to slide on two parallel conducting bars a in Fig. 2- Two resistors R_1 = 2.00 Ohm and R_2 = 5.00 Ohm are connected across the ends o t form a loop. A constant magnetic field B = 2.50 T is directed perpendicularly into the page. An external agent pulls the rod to the left with a constant speed of v - 8.00 m/s. Find a in both resistors, the total power delivered to the resistance of the circuit, the magnitude of the applied force that is needed to move the rod with this constant velocity, and done by the external agent within one minute?

Explanation / Answer

induced emf = B*v*l

applying kirchoff law for left loop

-I1*R1 + Bvl = 0

I1 = B*v*l/R1 = 2.5*8*0.35/2 = 3.5 A

for right loop

-I2*R2 + B*v*l = 0

I2 = B*v*l /R2 = 2.5*8*0.35/2 = 3.5 A

current through rod I = I2 + I3 = 7 A

F = I*L*B

F = 7*0.35*2.5 = 6.125 N

(d)

Wdone = P*t = F*v*t = 6.125*8*60 = 2940 J

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