A and B are parallel long straight wires in a vacuum 2.3cm apart. Wire A and wire B carry an equal current I. There's also a diagram which looks like this: http://puu.sh/luD5x/a902f526e5.png (a)...

A and B are parallel long straight wires in a vacuum 2.3cm apart. Wire A and wire B carry an equal current I.

There's also a diagram which looks like this:

http://puu.sh/luD5x/a902f526e5.png

(a) Explain why wire a exerts A force on wire B.

(b) In which direction is this force on the wire B? Explain briefly how you obtained your answer.

(c)Calculate the current I given that both wires experience a force of 3.2*10^-9 N per unit of lenght.

B=`alpha` I/2 , for a long straight wire.` `

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ishpiro | College Teacher | (Level 1) Educator

Posted on

a) The wire A exerts a force on wire B because electric current in wire A creates magnetic field around the wire. Since there is electric current (moving electric charges) in wire B as well, the magnetic field exerts force on these charges, which results in force on wire B.

 

b) The direction of this force is downward.

This could be determined by applying the right-hand rule. First, determine the direction of the magnetic field of wire A. If you curl the fingers of your right hand in a way that you thumb points in the same direction as the current, you will find that at the location of wire B, the magnetic field is pointed out of the page and towards you.

To find the direction of the force from this field on wire B, point the palm of your right hand along the current in wire B, and curl your four fingers towards the direction of the magnetic field (towards you.) Your thumb will then point downward - this is the direction of the force.

You can also remember the shortcut that the straight wires with the currents in the same direction will attract. This is in contrast to the electrostatics, where the like charges repel.

c) The formula for the force per unit length from one wire on another is

` `

and the magnetic field is

` `

Combining the two expressions results in

`f = (mu_0i^2)/(2pir) `

Since the force is known, we can solve for the current:

`i = sqrt((2pirf)/(mu_0))`

Plugging in all the given values (convert the distance between the two wires into meters, r = 0.0023 meters), and using `mu_0 = 4pi*10^(-7) (T*m)/A` ,we get

`i = 6*10^(-3) A`

 

The current in both wires is 6 mA, if the wires experience the force of 3.2 N per unit length.

 

 

 

Sources:

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