Question

This is an electromagnetic challenge:

Answer 1

We have three infinite conductors: conductor #1 and conductor #3 are fixed in the space, and the currents i_1 and i_2 flow into each of them respectively. Conductor #2 is free to move in the space, and the current i_2 flows into it. Furthermore, all those three currents have the same orientation. Give the condition on i_1, i_2, i_3, d_12, and d_13 such that the conductor #2 is in equilibrium. Show your work.

Answer 2

@IrishBoy123

Answer 3

@rational

Answer 4

@Kainui

Answer 5

@MTALHAHASSAN2

Answer 6

oops.. it is d_23, not d_13

Answer 7

@Michele_Laino

if there is no reply to this, pls don't post a solution. i will definitely take this on, but i have a UK Bank Holiday to negotiate before I am allowed to plug my brain in again.

Answer 8

sorry a student asked for my help!

Answer 9

this is Biot Savart

the wires are constrained, they must live in the same plane, yes?

Answer 10

yes!

Answer 11

hint:
the resultant force acting on the wire #2 has to be a null vector

Answer 12

my hint, above it is the only help which I can give to the solver

Answer 13

hardly a hint!!! Lorentz force demands it

i'll post something later but i smell a rat, ie i would not be surprised to find there is a really easy symmetry to this that obviates the need to do any Biot Savart.....

Answer 14

@Nnesha

Answer 15

\[\frac{i_1}{d_{12}} = \frac{i_3}{d_{23}}\]
using \[B_1 = \frac{\mu_o I_1}{2 \pi R_1}; \ B_3 = \frac{\mu_o I_3}{2 \pi R_3}\]
for an infinite wire from Biot-Savart, and because
\[\vec B_1 = - |B_1| \hat z; \ \vec B_3 = |B_3| \hat z; \ \vec B_1 + \vec B_3 = 0 \]
where
\[\hat z\] points upwards from the page,

all ultimately from Maxwell's Equations

Answer 16

i2 is irrelevant

Answer 17

good job! @IrishBoy123