A circular loop of diameter 10 cm, carrying a current of 0.20 A, is placed inside a magnetic field = 0.30 T
) The normal to the loop is parallel to a unit vector
= -0.60 î - 0.80 ĵ. Calculate the magnitude of the torque on the loop due to the magnetic field.

Question

A circular loop of diameter 10 cm, carrying a current of 0.20 A, is placed inside a magnetic field  = 0.30 T 
) The normal to the loop is parallel to a unit vector 
= -0.60 î - 0.80 ĵ. Calculate the magnitude of the torque on the loop due to the magnetic field.

Quizplus · Accepted Answer

The torque on a current loop in a magnetic field is given by:
τ = NIAB sinθ
where N is the number of turns, I is the current, A is the area of the loop, B is the magnetic field, and θ is the angle between the normal to the loop and the magnetic field. Since the normal is parallel to -0.60 î - 0.80 ĵ, we know that θ is 90 degrees. Therefore, sinθ = 1.
Using the formula above, we have:
τ = (0.20 A)(π(0.05 m)²/4)(0.30 T)(1) = 4.7 × 10^-4 N ∙ m
Therefore, the answer is A.

A Circular Loop of Diameter 10 Cm, Carrying a Current