Respuesta :
Answer:
9.25 x 10^-4 Nm
Explanation:
number of turns, N = 8
major axis = 40 cm
semi major axis, a = 20 cm = 0.2 m
minor axis = 30 cm
semi minor axis, b = 15 cm = 0.15 m
current, i = 6.2 A
Magnetic field, B = 1.98 x 10^-4 T
Angle between the normal and the magnetic field is 90°.
Torque is given by
τ = N i A B SinФ
Where, A be the area of the coil.
Area of ellipse, A = π ab = 3.14 x 0.20 x 0.15 = 0.0942 m²
τ = 8 x 6.20 x 0.0942 x 1.98 x 10^-4 x Sin 90°
τ = 9.25 x 10^-4 Nm
thus, the torque is 9.25 x 10^-4 Nm.
The magnitude of the torque on the coil is 9.25 x 10⁻⁴ Nm
Torque:
According to the question we have the following data:
Number of turns of the coil N = 8
Semi-major axis of the ellipse a = 40/2 cm = 0.2 m
Semi-minor axis, b = 30/2 cm = 0.15 m
Current in the coil, i = 6.2 A
Magnetic field, B = 1.98 x 10⁻⁴ T
The angle between the normal and the magnetic field is 90°.
So the torque on the coil is given by:
τ = NiABsinθ
Now, the area of ellipse:
A = πab
A = 3.14 x 0.20 x 0.15 = 0.0942 m²
Thus,
τ = 8 x 6.20 x 0.0942 x 1.98 x 10⁻⁴ x sin 90°
τ = 9.25 x 10⁻⁴ Nm
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