Question

You have three parallel wires of length L-5 m and mass 0.50 kg arranged to form an isosceles triangle with the top angle being θ 4. 80° and a side length of a-10 cm. If a current of 200 amps is run through the bottom two wires, what is the necessary current needed to run in the top wire such that it levitates at equilibrium? (10 pts] a. If instead of two separate wires, we had a single wire running 400 amps of current, how far above this new bottom wire would the top wire be at equilibrium? Assume the top wire has the same current as you determined in part A of this problem. [10 pts] b.

Answer 1

(a) The forces that are acting on the top wire due to other two wires can be understood from the following figure.

Hence the two forces are acting symmetrically on the top wire.

So, the angle between them must be equal to the 80⁰, i.e, the top angle in the isosceles triangle.

We can calculate the results force as follows,

$\vec{F}=\vec{F}_1+\vec{F}_2 \\ \Rightarrow \vec{F}=F_1\,sin(\theta/2)\hat{x}+F_1\,cos(\theta/2)\hat{y}+F_2\,sin(\theta/2)(-\hat{x})+F_2\,cos(\theta/2)\hat{y} \\ \Rightarrow \vec{F}=2F_1\,cos(\theta/2)\hat{y} \,\,\, (\because \left |F_1 \right |=\left |F_2 \right | )$

In order to have levitation effect, this force must be equal and opposite to the gravity force.

Hence,

$\Rightarrow 2F_1\,cos(\theta/2) = mg \\$

$\Rightarrow F_1=\frac{mg}{2\,cos(\theta/2) }$

Now, let us substitute the formula for the repulsive force (F₁) between two parallel wires carrying currents in the same direction, as in the present picture.

$\Rightarrow F_1=\left (\frac {\mu_0 I_1I_3}{2\pi a} \right ) L$

where, $\mu_0 = 4\pi \times 10^{-7} \text{ }\rm{H/m}$ is the permeability constant

I₁ is the current carried by W1 wire and is given as I₁=200 A

I₃ is the current carried by W3 wire, this need to be calculated.

a is the separation between these two wires, which is given as, a=10 cm = 0.1 m

L is the wire length, = 5 m

Along with these values, the mass value of the top wire is given as m=0.5 kg,

Let us substitute all these values, to find out the I₃ value.

$\Rightarrow \left (\frac {\mu_0 I_1I_3}{2\pi a} \right ) L=\frac{mg}{2\,cos(\theta/2) } \Rightarrow I_3=\left (\frac{mg}{2\,cos(\theta/2) } \right )\left (\frac {2\pi a }{\mu_0 I_1L} \right )$

$\Rightarrow I_3=\left (\frac{(0.5 \text{ kg}) (9.8 \text{ }\rm{m/s^2})}{2\,cos(80^0/2) } \right )\left (\frac {2\pi (0.1\text{ m}) }{(4\pi \times 10^{-7} \text{ }\rm{H/m}) (200 \text{ A})(5 \text{ m})} \right )\approx 1599\text{ A}$

Hence the current that is needed to flow in the top wire is about 1599 amp to levitate from the bottom two wires.

(b) In this case, we now know that the current in the top wire is 1599 A.

And if the bottom two wires are replaced by single wire that is carrying about 400 A current.

In these conditions, we need to find out the the distance that they both has to separated to levitate the top wire from the bottom wire.

In this case, the total magnetic force (F) that is balanced the weight (mg) as shown in the above figure to levitate the top wire.

Hence,

$\Rightarrow F=\left (\frac {\mu_0 I_1I_2}{2\pi a} \right ) L=mg\Rightarrow a=\left (\frac {\mu_0 I_1I_2 L}{2\pi mg} \right )$

By substituting the values, we can find out the separation distance as follows,

$\Rightarrow a=\left (\frac {\mu_0 I_1I_2 L}{2\pi mg} \right ) \\ \Rightarrow a=\left (\frac {(4\pi \times 10^{-7} \text{ }\rm{H/m}) (400 \text{ A})(1599 \text{ A})(5 \text{ m})} {2\pi\times (0.5 \text{ kg}) (9.8 \text{ }\rm{m/s^2})} \right )\approx 0.131 \text{ m}=13.1 \text{ cm} \\$

Hence, the top wire must be separated by a distance of 13.1 cm away from the bottom wire to the direction exactly above it.

Comment back if you need any additional information from the figures enclosed in this answer.