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
Question
Answer:
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