A circular loop carrying a current of 1.0 A is oriented in a magnetic field of 0.35 T. The loop has an area of 0.24 m2 and is mounted on an axis, perpendicular to the magnetic field, which allows the loop to rotate. If the plane of the loop is oriented parallel to the field, what torque is created by the interaction of the loop current and the field?

Respuesta :

Answer:

0.084 Nm

Explanation:

B = Magnetic field = 0.35 T

I = Current = 1 A

A = Area of the loop = 0.24 m²

[tex]\theta[/tex] = Angle between the solenoid and wire

N = Number of turns per meter = 1

The torque is given by

[tex]\tau=BIANsin\theta\\\Rightarrow \tau=0.35\times 1\times 1\times 0.24\times sin 90\\\Rightarrow \tau=0.084\ Nm[/tex]

The torque created by the interaction of the loop current and the field is 0.084 Nm

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