question.. (1 Viewer)

[Ragerunner]

Friend wants to know how to do this....

"Four vertical wires A, B, C and D are arranged so that when viewed from above they form a square of side 2.00 X 10^-2m. Currents of 3.00 A flow downward in wires A and C, and currents of 4.00 A flow upwards in wires B and D. What is the resultant force on wire D?"

The answer is 5.65 X 10^-5 N away from D and B along BD..

how do you get it?

 

[Xayma]

Sounds like a motors and generators question so Ill check excel and get back to you. Anyway you find the force in each of the wires concerned to D, ie AD, BD, CD and then do a sum of vectors. Ill post working in a sec. Hmm I keep getting 7.636 x10^-4N.m^-1. Ill figure it out

 

[Wohzazz]

is this a HSC question? oh no, i finished motor and gen and i can't do it
what happens exactly when there is 4 wires, the books only tell you what to do when there two

 

[Xayma]

When there are 4 your going to have to a sumation of vectors. After doing the forces in the parallell wires.

 

[Constip8edSkunk]

are there a diagram? i can't picture the situation... how does it form a square from the top view when the wires are vertical?

 

[Xayma]

I think the wires form the points of a square as in

A B

D C

 

[Wohzazz]

Originally posted by Xayma
When there are 4 your going to have to a sumation of vectors. After doing the forces in the parallell wires.

how do you do sumation of vectors?

 

[Xayma]

its the addition of the vectors, like done in projectile motion Ive just got sumation stuck in my head (is that even a word?)

 

[Ragerunner]

I tried using the F=kIIL/d but......that doesn't work.

Pretty dodgey question

 

[Constip8edSkunk]

i can see addition of vectors if we know the length of the wires....

 

[Ragerunner]

"form a square of side 2.00 X 10^-2m"

The length of the wire is 0.02 metres.

 

[Constip8edSkunk]

isnt 0.02 m the distance between the wires?

 

[Ragerunner]

But they are arranged to form a square.

I assume the distance apart and the length of the wire is the same...

 

[Constip8edSkunk]

See Diagram

edit: each vector can be found by F=kIIl/d, except l isnt given

 

[Constip8edSkunk]

the vecors are the attraction and repulsion between pairs of wires, which can b found by F/l =kII/d. ie. opposite wires attract and adjacent wires repel. but to use the formula, we need to know l, which isnt given although as you can see from the diagram, you can work out the correct direction

maybe theres another way...

 

[Constip8edSkunk]

the question said the wires were vertical though... which is why i was confused in the beginng.

 

[Ragerunner]

Ah yes, didn't read it properly.

Hmm....

 

[wogboy]

but to use the formula, we need to know l

Yes you need to know the length of the wire D in order to work out the force acting on it, there's a missing piece of information from the question.

Still assume the length of wire is L.

The force on D due to A is:
F1 = k*I_a*I_d*L / r = 2*10^-7 * 3 * 4 * L / 0.02 N
= 1.2 * 10^-4 * L N (directly away from A)

Likewise the force on D due to C is:
F2 = 1.2 * 10^-4 * L N (directly away from C)

The force on D due to B is:
F3 = k*I_b*I_d * L / r = 2*10^-7 * 4 * 4 * L / (0.02 * sqrt[2]) N
~ 1.13 * 10^-4 * L N (directly towards B)

Add F1, F2, and F3 as vectors (use the triangle technique), and the resultant force you get is:

5.67 * 10^-5 * L Newtons (away from B)

which closely corresponds to the given answer 5.65 * 10^-5 N. So the missing piece of info was probably that the wire D was 1m long.

[Edit] Corrected an error.

 

[Ragerunner]

Woo hoo thanks!

Makes more sense now

 

[Constip8edSkunk]

Actually i think they want the answer as Newtons per meter rather than just Newtons:

F/l = kI_1I_2/d

F/l _{vector_BD} = 2*10^-7 * 16 / (0.02*sqrt2) ....as teh distance between B and D = diagonal of the square, which can b found using pythagoras
F/l _{vector_CD} and F/l _{vector_AD} cancels out vertically,
thier horizontal components = (2*10^-7 * 12 / 0.02)*cos (pi/4) ....using trig

adding F/l _{vector_BD}, F/l _{vector_CD} and F/l _{vector_AD}:
Net F/l = [(2*10^-7 * 12 / 0.02)*cos (pi/4)]*2 - 2*10^-7 * 16 / (0.02*sqrt2) away from B,
=5.66 * 10^-5 Nm^-1 (3sf.)


edit: didnt see wogboy's post.... takes so long to post using subscript tags ...hehe