Physics - maximum torque?

Question

A 0.80 m length of wire is formed into a single-turn, square loop in which there is a current of 28 A. The loop is placed in a magnetic field of 0.3 T. What is the maximum torque that the loop can experience?

Answer 1

You have to keep stress on the term “maximum torque”. The torque faced by a current carrying coil in the magnetic field is

Tq (vector) = r x I (L x B) = r x I (L x B)
= I B x (r x L) by vector property = I * B x A
(where A is cross-section area vector, B mag. field, I current )

= I A B {sin (p)}

The torque is maximum when B (magnetic field) is perpendicular to the area vector (p =90 degree) (note: if floor is the coil then a vector from floor to roof is area vector. In other words, if maximum lines of magnetic force pass through coil (p=0) so torque will be zero. Here, for maximum torque [sin (p) = 1]

Also coil is made of wire of length (L = 0.80 m) thus each side of square coil (x=L/4=0.20). The area of coil A = 0.2*0.2 = 0.04 m^2

(Tq) max = I A B = 28 * 0.04 * 0.3 = 0.336 Newton-meter

Answer 2

Torque = Force x distance from rotation axis x sintheta Force = B x I x L for the long side of the coil so torque = B x I x L x r x sin(90) = B x Area of coil x 1 = 0.32 x 1.4 x 1.4 for one coil. Just multiply by N for N coils.

Answer 3

T=(a^2) I B
T= torque
I current
B magnetic firld strength
a -one side og a square loop

T=(0.8/4)^2 x 28 x 0.3
T= 0.336 m N