# Force Between Two Parallel Current Carrying Conductor

Previously we have learned about the existence of a magnetic field that is due to a current-carrying conductor and the Biot – Savart’s law.

We again have also learned that an external magnetic field that generally exerts a force which is on a current-carrying conductor and the Lorentz force which is the formula that governs this principle.

Thus, from the two studies that we can say that any two current carrying conductors that when placed near each other will exert a magnetic force that is on each other. In this section, we will learn about this case which is in further detail.

### Force Between Two Parallel Current Carrying Conductors

We might not generally expect that the force which is between wires is used to define the ampere. We might also be surprised to learn that this force has to do something with why large circuit breakers burn up when they attempt to interrupt large currents.

The force which is between two long straight conductors and the conductors which are parallel as well and separated by a distance r can be found by applying what we have developed in preceding sections. In the figure, we can see the wires and their currents fields which they generally create and the subsequent forces they exert on one another. Let us now consider the field produced by wire 1 and the force it exerts on wire 2, that is we can call the force F2. The field which is due to I1 is at a distance which is r is given to be

B1=μ0I1/2πr

Figure 1. (a) which is mentioned above is the magnetic field that is generally produced by a long straight conductor perpendicular to a parallel conductor as indicated by RHR-2. Then in the figure that is (b) a view which is from above of the two wires that are shown in (a) with one magnetic field line which is shown for each wire. Here the term that is RHR-1 shows that the force which is between the parallel conductors is attractive when the currents are in the same direction. A similar analysis is shown in this topic that the force is repulsive which is between currents in opposite directions.

If we have three wires which are the parallel in the same plane as it is shown in Figure 2 which is with currents in the outer two running in opposite directions that is it possible for the middle wire to be repelled by both.

Figure 2. Which we can see is the three parallel coplanar wires with currents in the outer two in opposite directions.

3. Then if we suppose that the two long straight wires run perpendicular to one another without touching. We can say that does one exerts a net force on the other. If so, then we can ask what is its direction? And does one exert a net torque on the other? If so, what is its direction? We need to Justify our responses by using the right-hand rule.

4. The Use which is of the right-hand rule is to show that the force which is between the two loops in Figure 3 is said to be attractive

Figure 3. We can notice that the two loops of wire carrying currents can exert forces and torques on one another.

5. Electric field can be shielded by the Faraday cage effect.

Now let us consider the system which is shown in the figure above. Here we can see that we have two parallel current carrying conductors that are separated by a distance denoted by ‘d’, So here we can notice that that conductor 2 experiences the same magnetic field that is at every point which is along its length due to conductor 1. That is we can say that the direction of magnetic force is indicated in the figure and is found using the right-hand thumb rule. The direction of the magnetic field which downwards due to the first conductor.

From Ampere’s law of the circuital we can say that the magnitude of  the field due to the first conductor can be given by the following:

Ba=μ0I1/2πd

The force which is on a segment of length denoted by letter L of the conductor 2 due to the conductor 1 can be given as follows:

F21=I2LB1=(μ0I1I2/2πd)L

Similarly, we can calculate the force which is exerted by the conductor that is the 2 on the conductor 1. So we can see that conductor which is in the 1 experiences the same force that is generally due to conductor 2 but the direction which is in the opposite. Thus we can pen it down as follows:

F12 = F21