motor effect problem (1 Viewer)

[joesmith1975]

i cant figure out how to do this question between 3 wires X, Y, Z, 1 having 4 A{X}, 5A{Y}, 6A{Z} , also the current is flowing in the same direction. The distance between X and Y is .1 m and the distance between Y and Z is .15m.

Calculate the resultant force per metre on the wire X.

 

[echelon4]

here's my attempt:

let F1= force between X and Y
let F2= force between X andZ

using formula F/l=k(I1xI2)/d

F1=0.00004
F2=0.0000192

Ftotal=F1+ F2
=0.0000592N
The direction of the force would be X the wire moving towards Y


EDIT

forgot one thing. the force between y and z (assume it's F3)

so resultant would be F1+F2+F3=0.00004+0.0000192+0.00004
=0.0000992N

Is that right?

 

[helper]

The Force between Y and Z does not effect X so is not included in calculations.

 

[echelon4]

I don't quite understand why force YZ isn't considered. If Y is being pulled towards Z, and X is being pulled towards Y, shouldn't it be a addition of both the force between YZ and XY?

 

[helper]

You need to look at the forces acting on a particular body, when you calculate the net force. Since YZ is not acting on X you don't include it.

If you were given enough information to calculate the changing distance Y is from Z, then yes you could include an integration factor in the force YX to account for the changing distance, if there is actual movement. This is not needed for the HSC.

 

[Riviet]

The wires are not connected, and we are considering the force at any given time, so the force of YZ would have absolutely no effect on X.

 

[nickthebiscuit]

here's my attempt:

let F1= force between X and Y
let F2= force between X andZ

using formula F/l=k(I1xI2)/d

F1=0.00004
F2=0.0000192

Ftotal=F1+ F2
=0.0000592N
The direction of the force would be X the wire moving towards Y

That is the correct answer then, yes?

 

[Alfred_rulz]

Nickthebiscuit is correct.

The answer to this question is 5.92 x 10^-5 N