Question

A flat rectangular loop of wire carrying a 4.0-a current is placed in a uniform 0.60-t magnetic field. the magnitude of the torque acting on this loop when the plane of the loop makes a 30° angle with the field is measured to be 1.1 n ∙ m. what is the area of this loop?

Answer 1

Torque on a closed current carrying loop is given by formula

$T = NiABsin\theta$

here

N = number of turns = 1

i = current = 4 A

A = area of the loop

B = magnetic field = 0.60 T

$\theta$ = angle between normal of the plane and magnetic field = 90 - 30 = 60 degree

Now we will have

$1.1 = 1* 4 * A * 0.60* sin60$

$1.1 = 2.078*A$

$A = 0.53 m^2$

So area of the loop is 0.53 m^2