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Damm [24]
2 years ago
9

Suppose a straight 1.00-mm-diameter copper wire could just "float" horizontally in air because of the force due to the Earth’s m

agnetic field which is horizontal, perpendicular to the wire, and of magnitude 5.0x10-5 T. What current would the wire carry? Does the answer seem feasible? Explain briefly.
Physics
1 answer:
insens350 [35]2 years ago
8 0

To solve this problem it is necessary to apply the concepts related to the Force since Newton's second law, as well as the concept of Electromagnetic Force. The relationship of the two equations will allow us to find the magnetic field through the geometric relations of density and volume.

F_{mag}= BIL

Where,

B = Magnetic Field

I = Current

L = Length

<em>Note: F_{mag}  is a direct adaptation of the vector relation F=q \times V \times B</em>

From Newton's second law we know that the relation of Strength and weight is determined as

F_g = mg

Where,

m = Mass

g = Gravitational Acceleration

For there to be balance the two forces must be equal therefore

F_{mag} = F_g

BIL = mg

Our values are given as,

Diameter (d) = 1.0mm = 1*10^{-3}m

Radius (r) = \frac{d}{2} = \frac{1*10^{-3}}{2} = 0.5*10^{-3}m

Magnetic Field (B) = 5.0*10^{-5} T

From the relationship of density another way of expressing mass would be

\rho = \frac{m}{V} \rightarrow m = \rho V

At the same time the volume ratio for a cylinder (the shape of the wire) would be

V = \pi r^2 L \rightarrow L =Length, r= Radius

Replacing this two expression at our first equation we have that:

BIL = mg

BIL = ( \rho V)g

BIL = ( \rho \pi r^2 L)g

Re-arrange to find I

I = \frac{( \rho \pi r^2 L)g}{BL}

I = \frac{( \rho \pi r^2 )g}{B}

We have for definition that the Density of copper is 8.9*10^3 Kg/m^3, gravity acceleration is 9.8m/s^2 and the values of magnetic field (B) and the radius were previously given, then:

I = \frac{( (8.9*10^3 ) \pi (0.5*10^{-3})^2 )(9.8)}{5.0*10^{-5}}

I = 1370.05A

The current is too high to be transported which would make the case not feasible.

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