A straight, vertical wire carries a current of 2.60 A downward in...

Question

A straight, vertical wire carries a current of 2.60 A downward in a region between the poles of a large superconducting electromagnet, where the magnetic field has magnitude B = 0.588 T and is horizontal. What are the magnitude and direction of the magnetic force on a 1.00-cm section of the wire that is in this uniform magnetic field, if the magnetic field direction is (a) east?

Accepted Answer

Hey everyone. So this problem is working with magnetic forces. Let's see what it's asking us. We have a current carrying straight conductor placed inside of a solenoid perpendicular to its access. We have a current of a given magnitude, it's flowing in the positive y direction. The current is perpendicular to a uniform magnetic field of a given magnitude directed in the positive x direction. We are asked to calculate the magnitude and the direction of the magnetic force exerted on a two centimeter long segment of the conductor. So the first thing we can do here is recall that the magnetic force on a current carrying wires is given by F equals I L B times the sine of theta. And so if we take this term by term, we'll see that we actually have everything to solve this out. So our current is given to us as three amps, the length was given as two centimeters. I'm just gonna rewrite that as point oh two m, we stay in our standard units. The magnetic field magnitude was given to us as 1.5 tesla. And we know that the field and the current are perpendicular to each other. So that data is going to be 90 degrees. And so from there, it's just a plug in shot. So F is equal to three amps times .02 m Times 1.5 Tesla Times The Sine of 90. And that comes out to 0. mutants. And so when we look at our potential choices, we can eliminate A and B off of the bat. And so the next thing they're asking for is the direction of the magnetic force. And so for that, we need to recall our right hand rule. So for the right hand rule, if we point the pointer pointer finger in the direction of the current, which is the positive Y direction and our middle in the direction of the field, which is the positive X direction, then the thumb is pointing into the page. And so that is the correct answer for part two. And so when we look at our choices, answer C is the correct choice, it's .09 newtons going into the plane of the page, that's all we have for this one, we'll see you in the next video.

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