Question

A rectangular loop has 17 turns with sides w=.03m and l=.06m. The current is 9A. What is the force on each side and the torque on the loop if the external field is 0.5 T and is directed: (a) parallel to the plane of the loop (B1) (along the positive x-axis)? (b) normal to the plane of the loop (B2) (into the paper)?

Answer 1

Given :-

n = 17 turns

width = 0.3m

height = 0.6m

i = 9A

B = 0.5 T

Part a.)

A uniform magnetic field exerts no net force on a current loop but it does exert a net torque.The magnetic field is in the plane of the loop and is parallel to two sides.

The loop has a width w = 0.3m, a length l = 0.6m, and a current I = 9A & Magnetic field B = 0.5T , then the force on each of the left and right sides is F = IbB. The other sides experience no force because the field is parallel to those sides.

The two forces are in opposite directions so there's no net force but there is a torque that tends to make the loop rotate about an axis running through the center of the loop. The torque about this axis is:

τ = F (w/2) + F (w/2) = Fa = iwlB.

ab is the area of the loop, so the torque is, in this case,

τ = iAB, now considering the no of turns as n = 17 turns

we have

τ = n x iAB [ Where A - Area vector, B - magnetic field ]

= 17 x 9 x 0.3 x 0.6 x 0.5 Nm. = 13.77 Nm.

This is actually the maximum possible torque, when the field is in the plane of the loop.

Part b.)

When the field is perpendicular to the loop the torque is zero. In general, the torque is given by:

τ = niA × B [ Vector cross product ]

where the area vector is perpendicular to the plane of the loop. Its direction can be found from another right-hand rule: curl the fingers on your right hand in the direction of the current and your thumb, stuck out, points in the direction of the area vector.