- #1
JulienB
- 408
- 12
Homework Statement
Hi everybody! I am struggling with an old exam problem, could someone maybe help me out to figure it out? Here is how it goes:
A rod with resistance ##R = 0.1 \Omega## lays over two parallel tracks (resistance ##\approx 0 \Omega##, ##l=10##cm). A battery is connected between the tracks with a voltage ##U_0 = 15V##. The rod is attached with a massless string to a mass ##m=1##kg hanging down (see picture, the wheel is massless). A current ##I## goes through the closed circuit. A homogeneous magnetic field ##B_0## goes perpendicularly between the tracks. The acceleration due to gravity is to be taken as ##g =10##m/s##^2##.
1. Give an equation for the current ##I##.
2. Calculate the velocity at which the rod moves stationary.
3. Which fraction of the power delivered by the battery is transferred to mechanical work?
Questions have been translated from German, let me know if something is unclear.
Homework Equations
Magnetic flux: ##\phi_B = \int \vec{B} d\vec{A}##
Electromotive force: ##\epsilon = - \frac{d\phi_B}{dt}##
Lorentz force: ##\vec{F} = I \cdot \vec{L} \times \vec{B}##
Ohm's law: ##U = R \cdot I##
Power: ##P = \frac{U^2}{R} = \frac{dW}{dt}##
Work: ##W = F \cdot s##
The Attempt at a Solution
1. I will call the current induced by the battery ##I_0## and the current induced by the magnetic field ##I_{ind}##. First I determine the magnetic flux as a function of time:
##\phi_B = \int \vec{B} d\vec{A} = \vec{B} \vec{A}(t) = B_0 (A - \frac{1}{2} g \cdot l \cdot t^2)##
##\implies ## electromotive force: ##U_{\epsilon} = - \frac{d\phi}{dt} = -B_0 g l t##
With Ohm's law I get:
##I_{ind} = - \frac{B_0 g l t}{R}##
which leads me to:
##I = |I_{ind}| + |I_0| = \frac{U_0 + B_0 g l t}{R}##
Now that's neat. It was fast and pretty much straightforward. But I have a problem with this answer: say at time ##t=0## the rod is not moving. Then the acceleration of the rod and the change in area would instantly be ##\frac{1}{2} g t^2##. But what about the Lorentz force? Isn't it acting on the rod in the opposite direction and therefore we would get another equation for the current? A much more complicated one actually. I've given it a try and was unable to get anywhere since everything becomes interdependent.
Am I overcomplicating the problem with this question? I just feel like this is all wrong if we neglect the Lorentz force...
2. Assuming my answer in a was correct, I now introduce the Lorentz force acting in the opposite direction of the gravity force. For the velocity to be constant, there must be no net force acting on the rod, that is:
##F_L = F_G \iff I \cdot \vec{l} \times \vec{B} = - m \cdot \vec{g} \iff I \cdot l \cdot B_0 = m \cdot g##
##\iff \frac{U_0 + B_0 g l t}{R} l \cdot B_0 = m \cdot g##
##\implies t = \frac{mgR - U_0 l B_0}{B_0^2 l^2 g}## (time elapsed until velocity is constant)
Now I must know the acceleration at that time:
##\sum F = F_G - F_L = m \cdot a##
##\implies a = g - \frac{U_0 + B_0 g l t}{m R} l B_0##
And the velocity turns out to be:
##v = a \cdot t = \Bigg(g - \frac{U_0 + B_0 g l t}{m R} l B_0 \Bigg) \Bigg( \frac{mgR - U_0 l B_0}{B_0^2 l^2 g} \Bigg)##
What do you guys think? It simplifies a little but not much really, so I prefer to leave it like that for now so that we can discuss the most important steps without having a calculation mistake ruin it all.
3. This question is tricky, not really sure how to interpret it. Nevertheless I came up with that answer:
The force component related to the battery is ##F_{U_0} = \frac{U_0 l B_0}{R}##.
The power delivered by the battery is ##P_{U_0} = U_0 \cdot I_0 = \frac{U_0^2}{R}##.
Since power is work/time, I guess I can express the power used in mechanical work as:
##P_{U_0, mech} = \frac{W_{U_0, mech}}{t} = \frac{F_{U_0} \cdot s}{t} = \frac{U_0 l B_0}{R} \cdot s##
Then I can take the ratio between the two powers, which leads me to:
##\frac{P_{U_0, mech}}{P_{U_0}} = \frac{U_0 l B_0}{R} \cdot s \frac{R}{U_0^2} = \frac{l \cdot B_0}{U_0} s##.
That's not bad, but what to do with the ##s## now? I believe I cannot insert ##\frac{1}{2} a t^2## in there since the acceleration is not constant. I could put ##v \cdot t## using my results from b, but that doesn't answer the question for ##t < t_{stationary}##... Any suggestion?Thanks a lot in advance for your answers, I appreciate. Looking forward to read your suggestions.Julien.