Acceleration of a conducting bar on rails in a magnetic field

[nickmanc86]

1. In the arrangement shown, a conducting bar of negligible resistance slides along
horizontal, parallel, frictionless conducting rails connected as shown to a 4.0-Ω resistor.
A uniform 2.5-T magnetic field makes an angle of 30° with the plane of the paper. If
L = 80 cm and the mass of the bar is 0.40 kg, what is the magnitude of the acceleration of
the bar at an instant when its speed is 1.5 m/s?

http://labman.phys.utk.edu/phys222/modules/m5/images/bar4.gif



2.$F_{b} =I \ast L \ast B \ast sin(\theta)$ and $F_{app} =m \ast a$



3. I set the two equations equal to one another and then proceed to solve for a. In this new equation I substitute in $I= \frac{V}{R}$ and $V=B \ast L\ast v(i)$The resulting equation of a= $\frac{ B^{2}\ast L^{2} \ast v(i)}{R \ast m} \ast sin(\theta)$. I come out with an answer of 1.88 however the correct answer is apparently suppose to be 0.9.

 

[TSny]

$F_{b} =I \ast L \ast B \ast sin(\theta)$

Can you describe the meaning of θ in this equation? What is it's value for this problem?

Can you describe the direction of the force on the bar?

$V=B \ast L\ast v(i)$

This equation does not account for the angle that the magnetic field makes to the plane of the paper.

 

[nickmanc86]

θ in the first equation would be the angle between the I*L vector and the B vector or the bar and the magnetic field so it should be 30 degrees?

For the second equation that would lead me to believe that because this equation originates from $F_{b} = q \vec{v} \times \vec{B}$ it should come out to be F=qvBsin(θ) and that finally plugging into the last equation it should actually be $sin^{2}(\theta)$ ?

I will try this and see what it comes out to.

 

[TSny]

θ in the first equation would be the angle between the I*L vector and the B vector or the bar and the magnetic field so it should be 30 degrees?

OK. It seems to me that the question statement is ambiguous about the direction of B. Suppose we let the x-axis be oriented left to right, let the y-axis be parallel to the bar, and the z axis out of the page. Then it appears that you are taking the direction of B such that B has a negative z component, a positive y component, and no x component. Then θ would equal 30^o as you stated. Another possible orientation would be for B to have a negative z component, a positive or negative x component, and no y component. Then θ would equal 90^o. There are an infinite number of other possibilities for the orientation of B where the x and y components are both nonzero and θ will have other values.

But, you can show that all of these orientations lead to the same answer for the acceleration of the bar. So, your choice should be ok.

For the second equation that would lead me to believe that because this equation originates from $F_{b} = q \vec{v} \times \vec{B}$ it should come out to be F=qvBsin(θ) and that finally plugging into the last equation it should actually be $sin^{2}(\theta)$ ?

I'm not sure about your reasoning here, but you are correct that there should be a factor of sinθ in the voltage expression. For me, it's easier to apply Faraday's law of induction: V = rate of change of flux. The flux will incorporate the direction of B with respect to the plane of the loop.