# Determine the magnitude and direction of the current in the 5.00 Ω resistor.

Two parallel rails with negligible resistance are 13.5 cm apart and are connected by a 5.00  resistor. The circuit also contains two metal rods having resistances of R = 15.0 Ω and 15.0 Ω sliding along the rails. The rods are pulled away from the resistor at constant speeds 4.00 m/s and 2.00 m/s, respectively. A uniform magnetic field of magnitude 0.0100 T is applied perpendicular to the plane of the rails. Determine the magnitude and direction of the current in the 5.00 Ω resistor.
µA

asked Feb 28, 2012 in Physics

This problem can be solved using Kirchhoff's Voltage Law, which states that the total voltage in a loop must sum up to 0.

In this diagram we have two loops, one on the left and the one on the right

Here is a diagram:

KVL at the left loops is:

+Bdv2 -I2*R2 - I1*R1 = 0

KVL at right loop:

+Bdv3 - I3*R3 + I1*R1 = 0

And at the junction we have:

I2 = I1 + I3

Now replace I2 with I1 + I3, in the KVL at left loop and get:

Bdv2 - (I1 + I3)*R2 - I1*R1 = 0

Solving for I3 we get:

I3 = (Bdv3 / R3)  +  (I1R1 / R3)

So, Bdv2-I1(R1+R2) - (Bdv3R2 / R3) - (I1R1R2 / R3) = 0

solving for I1 we get:

I1 = Bd(v2R3 - v3R2) / (R1R2 + R1R3 + R2R3)  upward.

Now just subtitute in the values given to get find I1, the current in the resistor.

d = 13.5 cm = 0.135 m

B = 0.01 T

v2 = 4 m/s

v3 = 2 m/2

R1 = 5 Ω

R2 = 15 Ω

R3 = 15 Ω

Therefore I1 = 1.08*10^-4 A  = 108 µA    upward

answered Mar 4, 2012 by ~Expert~ (1,318 points)
edited Mar 4, 2012 by Roar3215