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 19-cmcm-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 x 10³ N/m torque?

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AFOKE88 AFOKE88 · Answer 1 · 2020-03-06T10:42:43+01:00

Answer:

Current needed = 704A

Explanation:

Using the fomula; torque(τ) = (I)(A)(B)Sinθ

Where B = uniform magnetic field

I = current and A = Area

Diameter = 19cm = 0.19m so, radius = 0.19/2 = 0.095m

Area(A) = πr^(2) = πr^(2)

= π(0.095)^(2) = 0.0284 m^(2)

Now, B(earth)= 5x10^-5 T

While, we can ignore the angle because it's insignificant since the angle of the wire is oriented for maximum torque in the earth's field.

Now, if we arrange the formula to solve for charge (I):

I = (τ)/(A)(B)

I = (1.0x10^-3) / (0.0284)(5x10^-5)

I = 704A