Magnetic forces on current-carrying wires

[hooman]

Homework Statement

Three long straight thin conductors X, Y, Z lie on the same plane and are parallel to each
other as shown in Figure 1.Current flows in each conductor in the direction as shown. The
earth’s magnetic field may be neglected. The resultant magnetic force per unit length acting
on X is 0.50 mN m-1 , pointing towards X.
a) Is the direction of the current in Z shown in the diagram correct? Explain.
b) Determine the current in Z.

Relevant Equations

F/l= 0.50

please correct me if I'm wrong.

is it for the question b, F/l= 0.50 right? does that mean the resultant magnetic force at X is determine by the force at Y and Z? do I need to consider the direction to solve the equation? first step to know the direction of each conductors I use the right hand rule and I already get the direction of the forces for each conductors then I don't know whether I need to consider the direction like positive or negative to put in the equation. Please help me, thank you.

 

[Doc Al]

is it for the question b, F/l= 0.50 right?

That's given as the force per unit length acting on wire X. Although I am not sure what "pointing towards X" means. (What's the direction of the force on X? up, down, left, right, into the page, out of the page?)

does that mean the resultant magnetic force at X is determine by the force at Y and Z?

The resultant force on X is due to the magnetic fields from Y and Z that act on X.

 

[hooman]

That's given as the force per unit length acting on wire X. Although I am not sure what "pointing towards X" means. (What's the direction of the force on X? up, down, left, right, into the page, out of the page?)

The resultant force on X is due to the magnetic fields from Y and Z that act on X.

direction force of X is to the left

 

[Doc Al]

direction force of X is to the left

That's what I assumed they meant. (But that could have been stated more clearly.)

So, what's the force per length on X due to the current in Y?

 

[hooman]

That's what I assumed they meant. (But that could have been stated more clearly.)

So, what's the force per length on X due to the current in Y?

you mean the value or what?

 

[Doc Al]

you mean the value or what?

Yes. Magnitude and direction.

 

[hooman]

Yes. Magnitude and direction.

2.5x10^-4 , direction to the left

 

[Doc Al]

2.5x10^-4 , direction to the left

So what additional force must wire Z exert on X to produce the given resultant?

 

[kuruman]

The resultant magnetic force per unit length acting
on X is 0.50 mN m-1 , pointing towards X.

I find this statement confusing (not your fault). The magnetic force per unit length acting on wire X must be perpendicular to the wire marked X. As such, the force must point either away from wire Y (or Z) or towards it. "Towards X" makes no sense to me. Regardless of this, how do you propose to approach the problem if the direction of the force on X were given to you in a way that makes sense?

 

[hooman]

So what additional force must wire Z exert on X to produce the given resultant?

is it 0.49975 ?

 

[hooman]

I find this statement confusing (not your fault). The magnetic force per unit length acting on wire X must be perpendicular to the wire marked X. As such, the force must point either away from wire Y (or Z) or towards it. "Towards X" makes no sense to me. Regardless of this, how do you propose to approach the problem if the direction of the force on X were given to you in a way that makes sense?

i would consider the direction of force is away from Y

 

[Doc Al]

is it 0.49975 ?

Note that the resultant (total) force/length on X is given as 0.5 mN/m = 0.5 10^(-3) N/m.

 

[Doc Al]

I find this statement confusing (not your fault). The magnetic force per unit length acting on wire X must be perpendicular to the wire marked X. As such, the force must point either away from wire Y (or Z) or towards it. "Towards X" makes no sense to me. Regardless of this, how do you propose to approach the problem if the direction of the force on X were given to you in a way that makes sense?

I agree with your comments about the confusing wording of the direction of the force. I wonder what book this is from? (The best I could assume is that they meant from Y or Z toward X. Not clear at all.)

 

[kuruman]

I agree with your comments about the confusing wording of the direction of the force. I wonder what book this is from? (The best I could assume is that they meant from Y or Z toward X. Not clear at all.)

Another interpretation might be "towards the positive x-axis", i.e. to the right.

 

[Doc Al]

Another interpretation might be "towards the positive x-axis", i.e. to the right.

Indeed!

 

[hooman]

Note that the resultant (total) force/length on X is given as 0.5 mN/m = 0.5 10^(-3) N/m.

i see, i thought the m here is metre. so i mistaken here. that's why when i do the solution i find it weird

 

[hooman]

I agree with your comments about the confusing wording of the direction of the force. I wonder what book this is from? (The best I could assume is that they meant from Y or Z toward X. Not clear at all.)

this is my assignment's question

 

[hooman]

Note that the resultant (total) force/length on X is given as 0.5 mN/m = 0.5 10^(-3) N/m.

so that means that resultant F/l is the sum of F/l from Y and from Z right?

 

[Doc Al]

so that means that resultant F/l is the sum of F/l from Y and from Z right?

Correct.

 

[kuruman]

this is my assignment's question

Yes, it is that. However it's like giving you the equation $y = 3 + x$ and asking you to find $x$ given that the absolute value of $y$ is 5. There are two answers.

 

[hooman]

Correct.

the question a , it ask for the direction of current Z correct or not. how to know it?

 

[Doc Al]

the question a , it ask for the direction of current Z correct or not. how to know it?

If you know how much F/l wire Z must contribute, then you can figure out what direction its current must have.

 

[hooman]

If you know how much F/l wire Z must contribute, then you can figure out what direction its current must have.

the F/l wire Z is 2.5x10^-4 also. the direction is correct?

 

[hooman]

Yes, it is that. However it's like giving you the equation $y = 3 + x$ and asking you to find $x$ given that the absolute value of $y$ is 5. There are two answers.

the question a , it ask for the direction of current Z correct or not. how to know it?

 

[kuruman]

the question a , it ask for the direction of current Z correct or not. how to know it?

That is what it is asking. To know the direction of the current, you need to know the direction of the force on wire X. This direction could be to the left or to the right. The statement of the problem says "towards X" which is neither to the left nor to the right.

 

[MartinCarr]

All about working out the force from each wire on X, the you know which way the current should be..

 

[hooman]

That is what it is asking. To know the direction of the current, you need to know the direction of the force on wire X. This direction could be to the left or to the right. The statement of the problem says "towards X" which is neither to the left nor to the right.

do you think it make sense if I said the "towards X" in the statement means the direction of force on X by Z? because it is impossible for the force on X by Y to be towards X as the current is in different direction. what do you think?

 

[kuruman]

do you think it make sense if I said the "towards X" in the statement means the direction of force on X by Z? because it is impossible for the force on X by Y to be towards X as the current is in different direction. what do you think?

No, it does not mean the direction of force on X by Z. The problem states clearly that it is the resultant on X, that is the sum of forces by Y and Z.

 

[hooman]

All about working out the force from each wire on X, the you know which way the current should be..

so, you minus the Fy and Fz because it is in different direction? what do you mean by the 2 current must in opposite direction? which current that you mean? X and Y or X and Z?

 

[MartinCarr]

See the attached file on the above post..