A square loop (sides A , B , C , D = 11 cm; I = 11 A) lies in the vertical yz pl

Question

A square loop (sides A, B, C, D = 11 cm;

I = 11 A)

lies in the vertical yz plane. Looked at from the right, the current is clockwise. There is a uniform magnetic field of B = 0.9 T in the vertical z direction.

(a) What are the forces on each side? (Note: direction can be in the: x direction, -x direction, y direction, -y direction, z direction, -z direction, or the force is zero)

(b) What is the torque on the loop?
__________ Nm

side A      magnitude      ________ N      direction     A square loop (sides A, B, C, D = 11 cm; I = 11 A) lies in the vertical yz plane. Looked at from the right, the current is clockwise. There is a uniform magnetic field of B = 0.9 T in the vertical z direction. (a) What are the forces on each side? (Note: direction can be in the: x direction, -x direction, y direction, -y direction, z direction, -z direction, or the force is zero) side A magnitude ________ N direction side B magnitude ________ N direction side C magnitude ________ N direction side D magnitude ________ N direction (b) What is the torque on the loop? __________ Nm

Explanation / Answer

a)

The magnitude of force acting along the side A is given by F =Bil =(0.9)(11)(0.11) =1.089N i.e in the +z direction

The magnitude of force acting along the side B is given by F =Bil =(0.9)(11)(0.11) =1.089N i.e in the +y direction

The magnitude of force acting along the side C is given by F =Bil =(0.9)(11)(0.11) =1.089N i.e in the -z direction

The magnitude of force acting along the side D is given by F =Bil =(0.9)(11)(0.11) =1.089N i.e in the -y direction

b)

The Torque acting on the loop is given by

t =NiAB

Number of turns N =1

Current passing through the coil is i =11A

Each side of the square (a) =11cm then area of the square (A) =a2 =(11*10-2)2 =0.0121m2

The  uniform magnetic field of (B )= 0.9 T

Now substituting the values we get t =NIAB =(1)(11)(0.0121m2)(0.9T) =0.12N.m

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