Respuesta :
Answer:
A circular coil of wire of 200 turns and diameter 2.0 cm carries a current of 4.0 A. It is placed in a magnetic field of 0.70 T with the plane of the coil making an angle of 30° with the magnetic field. What is the magnetic torque on the coil?
Magnetic torque on the coil = 0.08796 N.m
Explanation:
Given:
A circular coil with N turns and I current along with its diameter d which makes an angle (\theta) .
Number of turns of the coil, N = 200
External magnetic field, B = 0.7 T
Current in the circular coil, I = 4 A
Diameter of the coil, d = 2 cm = 0.02 m
Angle between the coil and B , [tex](\theta)[/tex] = 30°
Formula to be used:
Magnetic torque in a circular coil:
⇒ [tex]\tau = NIABsin(\theta)[/tex]
⇒ [tex]\tau = NIB(\frac{\pi d^2}{4})sin(\theta)[/tex] ...Area of cross section = πd^2/4
⇒ Plugging the values.
⇒ [tex]\tau = 200\times 4\times 0.7\times \frac{\pi (0.02)^2}{4}\times sin(30)[/tex]
⇒ [tex]\tau = 200\times 4\times 0.7\times \frac{\pi (0.02)^2}{4}\times 0.5[/tex] ...sin(30)=0.5
⇒ [tex]\tau = 0.08796\ N.m[/tex]
So,
The magnetic torque on the coil = 0.08796 N.m
