People have proposed driving motors with the earth's magnetic field. This is possible in principle, but the small field means that unrealistically large currents are needed to produce noticeable torques. Suppose a 16-cm-diameter loop of wire is oriented for maximum torque in the earth's field.

What current would it need to carry in order to experience a very modest 1.0�10?3N?m torque?

Question

contestada

Respuesta :

Manetho Manetho · Answer 1 · 2019-12-09T11:13:47+01:00

Answer:

I_max= 1000 A

Explanation:

Expression of torque is given by

τ= BIA Sinθ

B= strength of magnetic field

I= current

A= Area

for maximum torque θ=90°⇒sin 90=1

Now for I_max= τ/BA

Let us take the average magnetic field of the earth as 50μT.

Area is calculated as

[tex]A=\pi d^2/4=\pi\times0.16^2/4[/tex]

[tex]A=0.02 m^2[/tex]

Substitute the values and calculate for maximum current

[tex]I_{max}= \frac{\tau}{BA}[/tex]

[tex]I_{max}= \frac{1\times10^{-3}{50\times10^{-6}\times0.02}[/tex]

I_max= 1000 A