Question

The parallel rails with negligible are 14.5 cm apart and are connected by a resistor of resistance R_3 = 5.00 Ohm. The circuit also contains two metal rods having resistances of R_1 = 13.5 Ohm and R_2 = 15.0 Ohm sliding along the rails (see figure). The rods are pulled away from the resistor at constant speeds of V_1 = 4.00m/s and V_2 = 2.00m/s, respectively. A uniform magnetic field of magnitude B = 0.0100 T is applied perpendicular to the plane of the rails. Determine the current in R_3.

Answer 1

flux 1 = B*A1

A1 = area = l*dx

l = 0.145 m

induced emf 1 = B*A1/dt = B*l*dx/dt = B*l*v1 = 0.01*0.145*4 = 0.0058 v

flux 2 = B*A2

A2 = area = l*dx

l = 0.145 m

induced emf 2 = B*A2/dt = B*l*dx/dt = B*l*v2 = 0.01*0.145*2 = 0.0029 v

for left loop

E1 - I1*R1 - I3*R3 = 0

0.0058 - I1*13.5 - I3*5 = 0.........(1)

for right loop

E2 - I2*R2 + I3*R3 = 0

0.0029 - (I2*15) + (I3*5) = 0.......(2)

I1 = I2 + I3 ....(3)

I1 = 378 uA

I2 = 240 uA

I3 = 138 uA <<<<_---------answer