Magnetism and Current Carrying wires (?)

[mike412]

Magnetism and Current Carrying wires (?)

I have a homework problem my teacher gave and warned us it was "Very difficult". I've tried drawing out the forces and i think i have made some headway but the vector-nature of magnetism confuses me... For example, i see in some solutions the forces along the y cancel and i don't know how that conclusion is made (i might be using RHR wrong).

Anyway here is my question: (OR SEE ATTACHMENT)

Three current carrying wires are configured into a PENDULUM. With the top vertex being stationary and carrying current I2= 2A (OUT OF THE PAGE). Off either side of the vertex are wires having I1= sqrt((mg*pi)/micr) a length of 1m, mass m and the string length is 1/6 m.

Find the ANGLE that the wires come to equilibrium.



Here is my work:
** Forces on a Pendulum **
Fy= Tension (along Y) = Gravitational force (mg)
Fx= Tension (along x) = Magnetic Force (BIL sin (theta))

therefore: F= mg (sin/cos theta) and since F= BIL sin (theta)

BIL sin (theta)= mg (sin/cos theta) [sin theta cancels leaving]

*** theta= cos^-1 (mg/BIL) ***

B can be substituted with (micro0*I/ 2*pi*r) but as long as i can't figure out B i can't solve the problem. HELP!

Thanks for your time,
Mike

 

[Andrew Mason]

BIL sin (theta)= mg (sin/cos theta) [sin theta cancels leaving]

*** theta= cos^-1 (mg/BIL) ***

B can be substituted with (micro0*I/ 2*pi*r) but as long as i can't figure out B i can't solve the problem. HELP!

Express the horizontal force on each wire as a function of the Tension (what forces contribute to the tension in the string?) and the repulsive forces between the wires. Use $B = \mu_0 I /2\pi r$ as the B from the other wires (where r is the distance to the centre of the other wire). Be careful. There are two wires providing the horizontal magnetic force on each of the two hanging wires.

AM

 

[berkeman]

The B fields will be different at the two pendulum wire locations. Use the right-hand rule to draw the two vector summations for the B field at each of the two pendulum wires, and keep track of the angles as variables. Note that the two pendulum wires will not necessarily be at the same angles out from the vertical when they settle out...

BTW, I don't get where in the text of the question they say that the strings to the two pendulum wires are 1/6 m long, and in the figure they are labeled as 1m long. Do you understand what they are saying?

 

[mike412]

This is what i have as far as forces:

For the current coming out of the page: (Bottom right)

a) Y-axis forces: (B1)(I1)Lsin(theta) = mg

b) X-axis forces: (B1)(I1)Lcos(theta) = (B2)(I1)L


Then the lower left looks like:

a) Y-axis: Tsin(theta) = mg + (B1)(I1)Lsin(theta)

b) X-axis: Tcos(theta) = (B1)(I1)Lcos(theta) + (B2)(I1)L

Is this correct so far...?

then i can substitute...

B1 = (mu)(2A) / (2 pi)

B2 = (mu) (I1) / (2 pi r) where r is the distance between the two hanging wires (1 meter)

 

[mike412]

oh yeah, typo on the figure length is 1/6 m. also no mass is given i assume it cancels somewhere in the problem unless the answer is to have mass in it. (?)

 

[Andrew Mason]

This is what i have as far as forces:

For the current coming out of the page: (Bottom right)

a) Y-axis forces: (B1)(I1)Lsin(theta) = mg

b) X-axis forces: (B1)(I1)Lcos(theta) = (B2)(I1)L


Then the lower left looks like:

a) Y-axis: Tsin(theta) = mg + (B1)(I1)Lsin(theta)

b) X-axis: Tcos(theta) = (B1)(I1)Lcos(theta) + (B2)(I1)L

Is this correct so far...?

then i can substitute...

B1 = (mu)(2A) / (2 pi)

B2 = (mu) (I1) / (2 pi r) where r is the distance between the two hanging wires (1 meter)

What angle are you calling $\theta$? I would use theta as the angle between the string and the vertical. As was correctly pointed out by berkeman, the angles will not be the same because the forces are different, so it would be a good idea to use different symbols for the right and left angles.

What about the Tension in the string to the right wire?

Also, I think the diagram is wrong and the strings really are 1/6 m. and the wires are 1 m. long. And the mass of the wires has to be a factor because the electrical forces are independent of mass.

AM