contestada

Assume that the loop is initially positioned at θ=30∘ and the current flowing into the loop is 0.500 A . If the magnitude of the magnetic field is 0.300 T , what is τ, the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field?

Respuesta :

cjmejiab

To develop this exercise, it is necessary to apply the definitions given by Faraday in its laws for the calculation of solenoids.

By definition the electromagnetic Torque is described as

[tex]\tau = NIBAsin\theta[/tex]

Where,

N = Number of loops

I = Current

B = Magnetic Field

A = Cross-sectional Area

[tex]\theta =[/tex] Angle between the loop and the magnetic field.

At this case the angle would be

[tex]\theta= 90-30 = 60[/tex]

There is only one loop for this calculations, then our values are

I = 0.5A

B = 0.3 T

[tex]A = (0.04)(0.02) = 8*10^{-4}m[/tex]

N = 1

Replacing in our equation we have

[tex]\tau = NIBAsin\theta[/tex]

[tex]\tau = 0.5*0.3*(8*10^{-4})sin60[/tex]

[tex]\tau = 1.039*10^{-4}Nm[/tex]

Therefore the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field is [tex]1.039*10^{-4}Nm[/tex]

The net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field is 1.039×10⁻⁴ Nm.

What is magnetic field?

The magnetic field is the field in the space and around the magnet in which the magnetic field can be fill.

The magnetic field in terms of electromagnetic torque can be given as,

[tex]B=\dfrac{\tau}{NAI\sin\theta}[/tex]

Here, (τ) is the electromagnetic torque, (N) is the number of loops, and (I) is the current, (A) is the cross-section area and (θ) is the angle between the magnetic field and loop.

The loop is initially positioned at θ=30∘. For the vertical axis, the angle will be 60 degrees (90-30). The current flowing into the loop is 0.500 A.

The magnitude of the magnetic field is 0.300 T. The dimensions of the loop is 0.04×0.02 m.  Plug in the values in the above formula for 1 loop as,

[tex]0.3=\dfrac{\tau}{(1)(0.04\times0.02)(0.5)\sin(60)}\\\tau=1.039\times10^{-4}\rm\; Nm[/tex]

Thus, the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field is 1.039×10⁻⁴ Nm.

Learn more about magnetic field here;

https://brainly.com/question/7802337

Otras preguntas