Answer:
The torque on the loop is [tex]2.4 \times 10^{-2}[/tex] Nm
Explanation:
Given:
Current [tex]I = 1.6[/tex] A
Magnetic field [tex]B = 0.30[/tex] T
Area of loop [tex]A = 0.14[/tex] [tex]m^{2}[/tex]
Angle between magnetic field and area vector [tex]\theta =[/tex] 21°
Form the formula of torque in case of magnetic field,
г [tex]= MB \sin \theta[/tex]
Where [tex]M =[/tex] magnetic moment
[tex]M = IA[/tex]
г [tex]= IAB \sin 21[/tex]
г [tex]= 1.6 \times 0.30 \times 0.14 \times 0.3583[/tex]
г [tex]=2.4 \times 10^{-2}[/tex] Nm
Therefore, the torque on the loop is [tex]2.4 \times 10^{-2}[/tex] Nm
