Induced EMF in a Triangle: Determining EMF and External Force

AI Thread Summary
The discussion focuses on calculating the induced electromotive force (emf) in a circuit involving a conducting bar moving through a magnetic field. Using Faraday's Law and the Hall Effect, the induced emf is derived as emf = Bvxtan(θ), where θ is the angle of the triangle. To maintain the bar's uniform motion, an external force (F_app) must counteract the induced force, resulting in F_app = B^2vx^2tan^2(θ)/R. The signs of the equations are debated, with the consensus that the direction of current flow determines the sign, but a positive sign for F_app is preferred for clarity. The discussion concludes with the acknowledgment that both emf and F_app can be expressed as functions of time, given the varying dimensions of the triangle.
SataSata
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Homework Statement


Magnetic Field B is going into the plane.
Bar is moving to the right with velocity v.
Neglect resistance of the conducting bar and all contacts.
a.) Determine the induced emf in the circuit.
b.) To maintain the uniform motion of the conductive bar, there must be an external force Fapp to pull the conductive bar. How much should Fapp be?
upload_2016-4-1_16-37-58.png


Homework Equations


Let the length of conducting bar in the triangle be ##l##
Hall Effect: ##emf = Blv ##
Faraday: ##emf = -d\Phi/dt##
##\Phi = BA##
##A=xl/2##
##l=xtan\theta##
##F_B=IlB##

The Attempt at a Solution


Using Faraday's Law:
##emf = -d\Phi/dt = -BdA/dt##
##dA/dt = dA/dx \times dx/dt##
##dA/dt = vxtan\theta##
##\therefore emf = -Bvxtan\theta##
Or, Using Hall Effect:
##emf = Blv = Bvxtan\theta##
Next, ##F_app = -F_B##
## Current I = emf/R = -Bvxtan\theta/R##
##F_app= IlB = B^2vx^2tan^2\theta/R##

Can anybody check if my attempt is correct? Are all the negative signs correct?
Faraday's Law and Hall Effect give different signs. Which one is correct?
Fapp sign is supposed to be positive or negative?
 
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SataSata said:
Faraday's Law and Hall Effect give different signs. Which one is correct?
Depends on the unspecified direction of your current flow. You don't need that direction, however.
I would choose a positive sign for the force needed to pull the bar, which means the induction leads to a negative force. But that is just an aesthetic choice (have force and velocity with the same sign convention if v is positive), the physical direction of the force is given in the problem statement already.
 
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Thank you for your reply mfb.
But is it right to express my answer in terms of x? Since x is a function of time, I can simply write it as x(t) and so emf and Fapp will both be a function of time too. However, the question simply asked for emf induced and Fapp, which caused me to doubt my answers.
In a non-triangle shape like a rectangle that's extending in length, the breadth is a constant and hence I can write my answer with it. But in this case, all sides of the triangle are varying and so I'm not sure how to express my answers.
 
Well, both depend on time here. Expressing them as function of x or t should be fine.
 
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