A horizontal rectangular surface has dimensions 2.80 cm by 3.20 c...

Question

A horizontal rectangular surface has dimensions 2.80 cm by 3.20 cm and is in a uniform magnetic field that is directed at an angle of 30.0° above the horizontal. What must the magnitude of the magnetic field be to produce a flux of 3.10x10^-4 Wb through the surface?

Accepted Answer

Everyone. So this problem is working with magnetic flux. Let's see what they are giving us and what they're asking us. So we know that there is a square coil. They give us the length of the side. Um it is placed in a magnetic field. The plane of the coil makes an angle of 35° with the field lines. We know that there is a magnetic flux passing through the coil and we are asked to determine the magnetic field strength that generates this flux. So the first thing we're going to do is recall the um equation for the magnetic flux and so that is phi equals B. A. Times co sign of data. Now we're actually asked to solve for B. The magnetic field strength. And so this value up here is a little bit of a red herring. So we can just ignore that. So let's rearrange this in terms of magnetic field strength. So B equals by over a times cosine of theta. And we'll take this term by term. So we are given the flux five, that's two point oh five times 10 to the negative four Webbers area. Let's see it's a square coil. And they gave us a side. So we can recall that the area of a square is just the side length squared. And that comes out to 1.6 times 10 to the negative three m squared. And then theta. So theta is the angle between a perpendicular vector to the area and the magnetic field. Okay, so we have a magnetic field and they tell us that we have The plane of a coil that takes that this angle is 35°. But theta is actually The angle between the magnetic field and a vector that is perpendicular to the area. So Theta is going to be 90 -35, which is 55°. That's just a little tricky part to this problem. But from there it's just plug and chug. So we have two point oh five times 10 to the minus four Webbers over the area, to be calculated to be 1.6 times 10 to the minus three m squared times the cosign of 55. And we can solve that for our magnetic field strength of 0. Tesla. All right. And looking at our answers that aligns with B. So the correct answer for this problem is B. That's all we have for this one. We'll see you in the next video.

A horizontal rectangular surface has dimensions 2.80 cm by 3.20 cm and is in a uniform magnetic field that is directed at an angle of 30.0° above the horizontal. What must the magnitude of the magnetic field be to produce a flux of 3.10x10^-4 Wb through the surface?

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