A current of 4.0 A is maintained in a single circular loop having a circumference of 80 cm. An external magnetic field of 2.0 T is directed so that the angle between the field and the plane of the loop is 20°. Determine the magnitude of the torque exerted on the loop by the magnetic forces acting upon it. Group of answer choices 0.27 N ⋅ m 0.41 N ⋅ m 0.38 N ⋅ m 0.77 N ⋅ m 0.14 N ⋅ m

0.14 N.m

Explanation:

Applying,

T = BIAsinФ Equation 1

Where T = Torque, B = magnetic Field, I = current, A = Area of the circular loop, Ф = angle between the field and the plane loop.

But,

A = πr² Equation 2

And,

c = 2πr Equation 3

Where c = circumference of the circular loop, r = radius of the circular loop

make r  the subject of the equation in equation 3

r = c/2π Equation 4

Give: c = 80 cm = 0.8 m, π = 3.14.

Substitute into equation 4 to get r

r = 0.8/(2×3.14)

r = 0.127 m.

Substitute into equation 2 to get the area of the circular loop.

A = 3.14(0.127²)

A = 0.051 m².

Also Given: B = 2 T, I = 4 A, Ф = 20°

Substitute into equation 1

T = 2(4)(0.051)(sin20°)

T = 0.408(0.342)

T = 0.1395 N.m

T ≈ 0.14 N.m

d.7.5cm is your answer