Direction of a moving charage and energy state of a magnetic dipole?

[Violagirl]

Homework Statement

1. A wire of length L and carrying current I is bent into a square loop to form a magnetic dipole. This dipole is then oriented in a uniform magnetic field of magnitude B such that the dipole is in its lowest energy state. What is the value of the dipole's energy with this orientation?

A. 0 B. -ILB C. -IL²B D. -1/4IL²B E. -1/16IL²B


2. A negative point charge moves with a velocity v next to a long, straight wire carrying a current I upward (in the plane of the page. Because of the charge's motion, there exists a force on the charge to the left. The charge must be moving:

A. to the left B. to the right C. upward D. downward E. out of the page


These were a couple of the multiple choice questions from a quiz I had in my physics course the other day and I wanted to go over them with someone on here and see if my logic on them was correct or not so I'll know for the final later on.

Homework Equations

For question 1:

μ = IA

U = -μ x B cos θ

F = IL x B

For question C:

F = qv x B

F = IL x B

The Attempt at a Solution

For question 1, I believed that answer to be C. I wasn't sure how to take into account when we're told that the dipole is in its lowest energy state. But I was thinking that since we a square loop, that in taking μ = IA, that A would be equivalent to L² so we'd get μ = IL² from that. Then to incorporate energy state, since the equation for energy is U = -μ x B, that the final answer would be U = -IL² x B.

For question 2, we're told that the moving charge is negative but that the current moves in an upward direction. I was thinking that a moving charge with a given velocity would be equivalent to current if the charge is positive. Since we're told that the point charge is negative, I rationalized that it must be moving in an opposite direction to the current and hence, would be moving in a downward direction, so answer D.

Thanks so much!

 

[BruceW]

yeah, I agree with your answer to question 1. And your answer to question 2 looks right. Although I can't be sure, because I'm not certain about how to imagine where the negatively charged particle is (relative to the wire). On the paper, I guess the wire is drawn as a vertical line, and the current is moving towards the top of the page? And on the paper, where is the negatively charged particle drawn? Is it to the left of the wire or the right of the wire?

 

[rude man]

I agree with question 1.

Question 2:

Indeterminate! Depends on where the charge is located along the wire.

If it's to the right of the wire it's moving in the downward direction.
If it's to the left of the wire then it's moving upward.

 

[Violagirl]

Thank you both for your input! Sorry didn't draw out the picture, I've attached it to this post! It looks like it's to the right of the wire so since the force is to the left of the charge, it would be moving downward then?

 

[rude man]

Thank you both for your input! Sorry didn't draw out the picture, I've attached it to this post! It looks like it's to the right of the wire so since the force is to the left of the charge, it would be moving downward then?

Correct.
One way to confirm this:
F = qv x B
Using unit vectors i,j,k only, since we're interested in direction only, not magnitudes,
-i = (-1)ζ x -k
where ζ is the unknown unit vector for v
So -i = -ζ x -k
the solution for which is ζ = -j.