A single turn current loop carrying a current of 542 a is in...

A single-turn current loop, carrying a current of 5.42 A, is in the shape of a right triangle with sides 42.7, 127, and 134 cm. The loop is in a uniform magnetic field of magnitude 81.0 mT whose direction is parallel to the current in the 134 cm side of the loop. What is the magnitude of the magnetic force on (a) the 134 cm side, (b) the 42.7 cm side, and (c) the 127 cm side? (d) What is the magnitude of the net force on the loop?

Transcript

00:01 In this problem you're given a single loop in the shape of a right triangle.

00:04 You have a current flowing and don't worry if you're chosen during this problem you chose the current to go counterclockwise.

00:14 All's well because then b would be pointing in the opposite direction.

00:18 All will be fine.

00:18 So don't you're getting magnitudes here relative things.

00:22 So don't worry about it.

00:23 Just choose and be consistent with it and choose b to be parallel to i in the hypotenuse sign.

00:31 So we have all the sizes l1, l2, and l3 are here.

00:35 We know the magnetic field magnitude.

00:36 We know the magnetic field direction.

00:38 You know the current that's flowing in the loop.

00:40 I'll explain this angle later.

00:42 The loop is in the xy plane.

00:45 Z, positive z, is out of the screen.

00:48 Negative z is into the screen.

00:49 You might say what do i need that for? i don't want to get any magnitudes.

00:52 Well you'd be right if all you were being asked was parts a, b, and c.

00:57 But part d wants to know the net.

01:00 That means nets are not generated by adding magnitudes or vectors.

01:06 You have to look at components.

01:08 The only time magnitudes can be added up and give you the magnitude of the actual sum is if they're all in the same direction.

01:16 That's a very special case.

01:18 It's not normally going to be the case.

01:20 So we have to look at components.

01:23 Okay but we'll get to that in a little bit.

01:26 First formula when you have, when you have, when you're this, and b i should mention is a uniform field.

01:32 When you have a uniform magnetic field and you have a straight wire you can use this formula here.

01:43 Il cross b.

01:46 L is the length vector.

01:50 It's along the direction of the, it points in the direction of the current.

01:54 And its magnitude is the length of that segment.

01:59 The magnitude of a cross product ilb sine theta.

02:05 Where theta is the angle between the l vector and the b vector when they are tail to tail.

02:12 Now i should mention if it wasn't a straight wire or b was not uniform then you have to, this turns into the force and infinitesimal piece of that wire.

02:26 So those segment infinitesimal segment and you have to use integration.

02:30 Nothing special about that but if you're not used to that or if you're on a calculator based class then obviously you don't worry about that.

02:41 All right so part a.

02:44 Wants it for the hypotenuse l3 f3 il3 b sine.

02:57 Remember i said it's when they are tail to tail.

03:00 Here is l3 vector here is b vector not drawn to scale just being representative.

03:08 Tail to tail the angle zero degrees.

03:12 So there is no force on the hypotenuse l3 segment zero newtons.

03:21 Okay b wants it for segment one f1 il1 b sine theta one...