Magnetic field and current cause a force

[MathewsMD]

Hi,

For question 37 (attached), I found the length of hypotenuse to be ~0.21m by letting a² + b² = c² which is simple enough. Then I let F_b = i L x B and since L and B are perpendicular this gives a force with magnitude ~0.3 N. Now, I realize the question is asking to find the respective force on each leg of the right triangle, so you divide by 2 to get the answer of 0.15N. Is this process correct?

Now I may be mistaken here which is why I'm posting this question, but isn't the force field created uniform? Why do we end up dividing by 2? Doesn't everything in this field experience a force of 0.3 N?

Also, shouldn't I also be adding the force from the other sides of the triangle besides the hypotenuse since there's current in the other 2 legs that increase the net force experienced by each side of the triangular loop?

Any clarifications would be great!

 

[rude man]

Look at the statement of the problem carefully. It says the B field is UNIFORM across the triangle. That means that your only concern is the current in one of the equal legs. The other two legs and their current mean nothing.

Things would be very different if the B field were externally generated and left constant at 0.7T, and THEN the current would be turned on, greatly un-homogenizing the B field.

 

[MathewsMD]

Look at the statement of the problem carefully. It says the B field is UNIFORM across the triangle. That means that your only concern is the current in one of the equal legs. The other two legs and their current mean nothing.

Things would be very different if the B field were externally generated and left constant at 0.7T, and THEN the current would be turned on, greatly un-homogenizing the B field.

Thanks for the clarification. Aside from that, is my process correct?

 

[rude man]

Thanks for the clarification. Aside from that, is my process correct?

I doubt it, since for one thing you calculated the length of the hypotenuse which was totally immaterial.

All you need is to compute the force on one of the equal legs with the given B field.

 

[MathewsMD]

I doubt it, since for one thing you calculated the length of the hypotenuse which was totally immaterial.

All you need is to compute the force on one of the equal legs with the given B field.

Okay, the numbers seem to work out now, thank you. Just wondering, since the magnetic field is perpendicular to the hypotenuse with current flowing through it, doesn't it have any effect?

Once again, I may be completely wrong here but for the equation F = iL x B, isn't that the magnetic field produced by the current carrying wire, not the force it experiences?

 

[rude man]

It has plenty of effect on the hypotenuse but none on the answer. The question asks solely for the force on one of the equal lengths.

 

[rude man]

Once again, I may be completely wrong here but for the equation F = iL x B, isn't that the magnetic field produced by the current carrying wire, not the force it experiences?

Not if the problem states explicitly that the B field within the triangle is uniform. It's not your worry how it got there. Of course the hypotenuse current and the other two leg currents contribute to the B field, but that is not your worry. The B field is stated to be B and is uniform within the area of the triangle.