A wire loop, 2 by 4 meters, of negligible resistance, is in the plane of the page with its left end in a uniform .5 T magnetic field directed into the page (The field is zero outside this region). A 5-ohm resistor is connected between points X and Y. The loop is being pulled to the right with a constant velocity of 3 m/s. Make all determinations for the time that the left end of the loop is still in the field, and points X and Y are not in the field.
a. Determine the potential difference between points X and Y.
b. On the figure show the direction of the current induced in the resistor.
c. Determine the force required to keep the loop moving at 3 m/s.
d. Determine the rate at which work must be done to keep the loop moving at 3 m/s.
E = -Blv
V = IR
F = qvB = qE = qV/r
F = P/v
F = BILsin*
I = BAcos*
The Attempt at a Solution
a. I said E = -Blv (with E being the emv and potential difference).
So -(.5 T)(2 m)(3/ms) = -3V
Question: Is my length correct? I believe I would use the distance between X and Y if it wants the potential between it...
b. I decided the direction to be counterclockwise (of the current). If the wire loop is being pulled to the right, the area to the right is increasing and the magnetic field to the left is decreasing. If the magnetic field is decreasing to the left, it is increasing to the right. According to Lenz's law, I take the opposite of this to find the current - which would mean counterclockwise direction - and hence the "-" sign in the equation -Blv.
c. I know that F=BILsin* and that I = V/R
So F = B(V/R)L (sin* isn't needed in this equation as the angles are 90)
I also know that E = -Blv (E = emf = V), so:
F = B((-BLv)/R)L, or nicely, -(B^2)(L^2)v/R
Thus, (.5 T)^2 x (4 m)^2 x 3 m/s / 5 ohms = -2.4 N
Question: I have to include the "-" sign in the Blv, right? I would think so since force is a vector and has direction, but I dont know why the force would be negative..
d. The rate at which work is done is power, and P = Fv
So I just multiple my answer in c by 3 m/s.
P = (2.4 N)(3 m/s) = 7.2 W (watts)
Again, I'm not sure of direction. I don't believe you can have "negative" power to the best of my knowledge.
Any help is greatly appreciated. I don't necessarily want answers because if I do the work I understand it better, but any direction to my faults would be a big help. Thanks, and have a great day.