To solve this problem it is necessary to use the concepts related to Torque generated by the current.
From its definition as magnitude, it can be described as
[tex]\tau = IAB sin\theta[/tex]
Where
A = Cross Sectional Area
I = Current
B = Magnetic Field
[tex]\theta =[/tex]Angle between the magnetic field lines and normal to the loop area
Our values are given as
[tex]\theta = 30\°\\I = 10A\\A = \pi r^2 = \pi * (5*10^{-2}) ^2 \\A = 7.85*10^{-3}m^2\\B = 12T[/tex]
Replacing at the previous equation
[tex]\tau = IAB sin\theta[/tex]
[tex]\tau = (10)(7.85*10^{-3})(12) sin30[/tex]
[tex]\tau =0.471 Nm[/tex]
Therefore the correct answer is 0.47Nm.
