Question

https://dl.dropbox.com/u/63664351/Physics/ELEC1.PNG
Figure above shows a very long horizontal wire PQ carrying a current of 100 A flowing in the direction from Q to P. A copper wire which has a diameter of 0.4mm and as long as the wire PQ is suspended horizontally at a distance 0.2 m below the wire PQ by two strings. If the density of copper is 8900 kgm-3, determine the direction and the minimum value of the electric current which must flow in the wire RS so that the tension in the strings become zero.

Answer 1

Magnetic F PQ- Magnetic F RS = 0 may be?

Answer 2

Find the formula of a magnetic field (B) from a current carrying wire from amperes law (or just look it up). Then recall the force on a current carrying wire due to a magnetic field F=iLB, where L is a given length, and B is from the part above and I is current in the bottom wire that you seek. You don't have the length so solve it as force per unit length. Use newtons F=Ma as the force you need from above where the mass will be a unit length and it can be calculated from the cross sectional area of the wire and its mass density. This is a good problem to work through, it hits on many aspects of physics. Give it a try and let me see what you get and I'll help some more.

Answer 3

@satellite73

Answer 4

@egenriether I can't find formula for B

Answer 5

\[B= \frac{ \mu I}{ 2pir}\]

Answer 6

I found http://academic.mu.edu/phys/matthysd/web004/l0222.htm
Force between parallel conductor. I want to know what is mean by F here. Its net force?

Answer 7

So I can just use this F/L to be equal to F/L from gravity? How come they are equal. One from current and one from gravity? Even their direction is different. One is radial and the other going down. If F is net (from current of F/L) should I just make this = 0? Net force =0 meant there is no tension right?

Answer 8

OOH I GET IT NOW.
From here: http://webphysics.davidson.edu/physlet_resources/bu_semester2/c14_forcewires.html

"Using the infinite wire equation, wire 1 sets up a magnetic field that wire 2 experiences. The magnitude of this field, at wire 2's location, is:

B1 = mo I1 / 2pd directed up.

To find the force on wire 2, use:

F = I2L ´ B1

We don't have a length to use for wire 2, but at least we can get the force per unit length:

F/L = I2 B1 = mo I1 I2 / 2pd"

SO this is cancelled by force per unit area is cancelled by gravity. But that assuming the F/L going up. But isn't the force caused by current either clockwise or anticlockwise?

Answer 9

weight of the 2nd wire is downward direction.
in order to get a Resultant Force on String or Tension = 0, a magnetic force from the current should be applying upwards.

So that Fb1-Fmg=0, however, the weight of wire 2 supported by 2 strings, therefore, Fb1-1/2 Fmg=0

Answer 10

anyone can provide comprehensive solution?

Answer 11

https://dl.dropbox.com/u/63664351/Physics/Openstudy.PNG

Answer 12

Ok, you've come pretty far on your own, good work! Here's how I solved it. (check my math I did it on the fly!)