Question

A 10.0-cm side square loop is fashioned from a single 4.00-m long, 100-g wire. One side of the loop is attached to a horizontal, frictionless axle. The loop is suspended in a region with a uniform 10.0-mT magnetic field oriented vertically. (a) What angle does the loop make with the vertical when it carries a current of 3.40 A? (b) What magnetic torque acts on the loop?

Please don't just copy the other tutors, majority of their answers are wrong.

Answer 1

A) The perimeter of the square loop = L*4 = 10*4 = 40cm = 0.4m

As the length of the wire is 4m so the square loop consists of 4/0.4 = 10 turns = N (say)

Magnetic force on the segment of wire opposite of the axle F₁= NBIL

• B = 10.0mT
• I = 3.40A
• L = 0.1m

So, this force acts horizontally towards axle

The adjacent sides, say, make an angle of with the horizontal.

The magnetic forces on them cancel each other.

Now all forces works through the center of those 3 sides

each arm weighs 100/4 = 25g = 0.025kg

Now momet of the gravitational and magnetic forces are equal in the equlibrium.

So,

The loop almost stays vertical.

B)

The total magnetic torque on the loop from previous equation=