The triangular loop of wire shown in the drawing carries a...

Name: The triangular loop of wire shown in the drawing carries a current of I = 4.06 A. A uniform magnetic field is directed parallel to side AB of the triangle and has a magnitude of 2.03 T. (a) Find the magnitude of the magnetic force exerted on each side of the triangle. (b) Determine the magnitude of the net force exerted on the triangle.
Uploaded: 2021-05-30T13:41:22-08:00
Duration: 7 min 49 s
Channel: Vishal Gupta
Description: The triangular loop of wire shown in the drawing carries a current of I = 4.06 A. A uniform magnetic field is directed parallel to side AB of the triangle and has a magnitude of 2.03 T. (a) Find the magnitude of the magnetic force exerted on each side of the triangle. (b) Determine the magnitude of the net force exerted on the triangle.

The triangular loop of wire shown in the drawing carries a current of I = 4.06 A. A uniform magnetic field is directed parallel to side AB of the triangle and has a magnitude of 2.03 T. (a) Find the magnitude of the magnetic force exerted on each side of the triangle. (b) Determine the magnitude of the net force exerted on the triangle.

Transcript

00:02 Hi, in the given problem here this is a wire loop in the form of a triangle like this.

00:15 And there is a magnetic field in a direction left to right.

00:24 This is a uniform magnetic field.

00:30 The current passing through this coil, the triangular loop is in a clockwise direction.

00:37 And its magnitude current carried by the loop is given as i is equal to 4 .70 ampere the right word uniform magnetic field is having a magnitude 1 .80 tesla.

01:05 Now in the first part of this problem we have to find the forces, experienced by all the three arms of this triangle the arm ab the arm b c and the arm ac here the length of this arm ab is given as 2 .00 meter and this angle is given as 55 degree so first of all using and here there is only one part of this problem we have to find all the forces along with the directions experienced by all these three branches.

01:48 So using the expression for the force in vector form it is f equals to il, the current element having a cross multiplication with the direction of magnetic field.

02:03 So this force is given as ilb sine theta where theta is angle between current element and the direction of magnetic field.

02:12 So first of all for the arm ab this angle is 180 degree and sine 180 degree we know is 0 therefore force experienced by the arm ab fab will be i lb into sine 180 degree which is zero so this force will come out to be 0 and this is first answer for this given problem.

02:48 Now for the force experienced by the arm ac, this arm ac is exactly perpendicular to the magnetic field so theta will be 90 degree.

03:04 Hence fac will be b -i -l into sine 90 degree.

03:14 Now, first of all, we should find the length of this arm.

03:24 The arm, and here this, here it will be, in place of length, it should be ac...

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