Question

Two rigid rods are oriented parallel to each other and to the ground. The rods carry the same current in the same direction. The length of each rod is 0.67 m, while the mass of each is 0.082 kg. One rod is held in place above the ground, and the other floats beneath it at a distance of 7.4 mm. Determine the current in the rods.

Answer 1

To develop this problem we will apply the concepts related to the Electromagnetic Force. The magnetic force can be defined as the product between the free space constant, the current (of each cable) and the length of these, on the perimeter of the cross section, in this case circular. Mathematically it can be expressed as,

$F= \frac{\mu_0}{2\pi} \frac{I^2L}{d}$

Here,

$\mu_0$ = Permeability free space

I = Current

L = Length

d= Distance between them

Our values are,

$L = 0.67m$

$m = 0.082kg$

$d = 7.4*10^{-3}m$

$I = \text{Current through each of the wires}$

Rearranging the previous equation to find the current,

$\frac{mg}{L} = 2*10^{-7} (\frac{I^2}{d})$

$I = \sqrt{\frac{mgd}{(2*10^{-7})L}}$

$I = \sqrt{\frac{(0.082)(9.8)(7.4*10^{-3})}{(2*10^{-7})(0.67)}}$

$I = \sqrt{44377.9}$

$I = 210.66A$

Therefore the current in the rods is 210.6A

Answer 2

Answer:

I = 298A

Explanation:

Given m = 0.082kg for both rods, r = distance between the rods = 7.4mm = 7.4 ×10-³m, L = 0.67m, μo = 4π×10-⁷Tm/A.

For the two rods to be in equilibrium, the sum of their weights must balance the magnetic force of attraction between them.

So

mg + mg = μo× I×I×L/(2πr)

2mg = μo×I²×L/(2πr)

Re arranging,

I = √(4π×mgr/μo×L)

I = √(4π×0.082×9.8×0.0074/(4π×10-⁷×0.67))

I = 298A