Magnitude of the Magnetic Field near a Circuit Inclined at an Angle

[Mark Zhu]

Homework Statement

Consider a circuit composed of two parallel metal tracks, a generator which produces a constant current i, and a straight wire, mass m, which can slide along the tracks. The coefficient of friction between the wire and the tracks is "mu" and the tracks are "l" apart. Suppose there were a constant, vertical "vector B" field. Relate the magnitude of "vector B" to the angle the tracks make with the horizontal, "theta," assuming that "theta" has the maximum value it can have without the wire starting to move.

Relevant Equations

Vector F = q * (vector v x vector B)
F_s = mu * n

For the front wire, I got the magnitude of the magnetic field in terms of the magnitude of the magnetic force, the current, "l," and the "theta". I am unsure how to proceed because I thought that the magnetic force is independent of any other forces. I am also just lost in general. Any help would be greatly appreciated. Thanks.

 

[TSny]

This is mainly a mechanics problem. Start with a carefully drawn free-body diagram for the wire that can slide.

Make sure you are visualizing the set-up correctly. The picture that was given is not great.

 

[Mark Zhu]

Ok, is the main idea that the magnetic force and friction force counter the gravitational force experienced by the wire, thus, keeping it stationary?

 

[TSny]

I think you have the right idea.

 

[Mark Zhu]

After separating the force on the wire into x and y components on the incline, I used the equation for the net force in the x direction and solved for the magnitude of the magnetic force:
B = (mg(sin(theta) - mu*cos(theta))/(i*l*cos(theta)*sin(theta))
According to the book, the numerator for this answer is correct, but the numerator is i*l*(cos(theta) + mu * sin(theta))
and I am not sure why. Any help would be appreciated.

 

[TSny]

?

After separating the force on the wire into x and y components on the incline, I used the equation for the net force in the x direction and solved for the magnitude of the magnetic force:
B = (mg(sin(theta) - mu*cos(theta))/(i*l*cos(theta)*sin(theta))
According to the book, the numerator for this answer is correct, but the numerator is i*l*(cos(theta) + mu * sin(theta))
and I am not sure why. Any help would be appreciated.

Mark, I would need to see the specific steps of your work in order to identify your error. Please specify the orientation of your x and y-axes. What forces act on the wire? What expressions did you get for the x and y components of each of these forces?

 

[Mark Zhu]

I have lined up my x-axis along the incline and the y-axis perpendicular to it, both in the plane of the page. Positive x direction is going down the incline going towards the right. For the net force on the wire in the x direction, my equation was:

m * g * sin(theta) - Mu * F_n - i * l * B * sin(theta)^2 = 0

Where
F_n = m * g * cos(theta)
is the normal force on the wire.

The first term in the equation is the x component of the gravitational force on the wire; the second term is the static frictional force; the third term is the magnetic force.

For the net force on the wire in the y direction, I just got that the magnetic force times cos(theta) equals 0.

I solved for B from the first equation.

I'm not sure where I went wrong.
Thanks a lot for helping.

 

[TSny]

m * g * sin(theta) - Mu * F_n - i * l * B * sin(theta)^2 = 0

sin²θ is not the correct factor for getting the x component of the magnetic force. What is the direction of the total magnetic force on the wire?

Where
F_n = m * g * cos(theta)

Both mg and F_mag will contribute to the normal force.

For the net force on the wire in the y direction, I just got that the magnetic force times cos(theta) equals 0.

If F_mag cosθ = 0, then wouldn't that imply that F_mag = 0?

[Note that there is a toolbar where you can find tools for superscripts, math symbols, etc.]

 

[Mark Zhu]

After your help, I got one right answer. However, there are two answers for B and I'm not sure why. I got the answer labeled in my book as (Ready to slide down). There is a second value for B with switched signs in the numerator and denominator that is labeled (Ready to slide up). I'm not sure what this second answer is.

 

[TSny]

The problem statement implies that you are considering a situation where the wire is on the verge of slipping. But it does not state whether the wire is just about to start sliding down the slope or up the slope. The answer is different for these two cases. Think about the direction of the friction force for each case.

 

[Mark Zhu]

I understand now. Thank you for all your time and help.