A square current loop 4.90 cm on each side carries a 510mA current. The loop is in a 1.40 T uniform magneticfield. The axis of the loop, perpendicular to the plane of theloop, is 30 degrees away from the field direction.
What is the magnitude of the torque on the current loop?

the torque on the current loop is 0.000857Nm

Explanation:

Given that                  

side of loop = 4.90 cm = 0.049 m

current on each side of loop =  510 m A = 0.51 A

magnetic field on the loop = 1.40 T

axis of the loop make an angle of = 30°

torque = ?

T = i x A x B x sin(30)

the torque on the current loop is 0.000857Nm

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow

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hope this