# Q. Find the torque and the force between two circular loops of wire, carrying the same currents I, and of the same radius R, when they are located a distance L apart, with L >> R, and with their axes parallel and the currents in the same direction. Express the torque and the force in terms of the angle q between their axes and their line of centers.

**Q.** Find the torque and the force between two circular loops of wire, carrying the same currents I, and of the same radius R, when they are located a distance L apart, with L >> R, and with their axes parallel and the currents in the same direction. Express the torque and the force in terms of the angle q between their axes and their line of centers.

## 1 Answers

The force between two long straight and parallel conductors separated by a distancercan be found by applying what we have developed in preceding sections. Figure 1shows the wires, their currents, the fields they create, and the subsequent forces they exert on one another. Let us consider the field produced by wire 1 and the force it exerts on wire 2 (call the forceF2). The field due toI1at a distanceris given to be

B1=μ0I12πrB1=μ0I12πr.[Figure a shows two parallel wires, both with currents going up. The magnetic field lines of the first wire are shown as concentric circles centered on wire 1 and in a plane perpendicular to the wires. The magnetic field is in the counter clockwise direction as viewed from above. Figure b shows a view from above and shows the current-carrying wires as two dots. Around wire one is a circle that represents a magnetic field line due to that wire. The magnetic field passes directly through wire two. The magnetic field is in the counter clockwise direction. The force on wire two is to the left, toward wire one.]Figure 1. (a) The magnetic field produced by a long straight conductor is perpendicular to a parallel conductor, as indicated by RHR-2. (b) A view from above of the two wires shown in (a), with one magnetic field line shown for each wire. RHR-1 shows that the force between the parallel conductors is attractive when the currents are in the same direction. A similar analysis shows that the force is repulsive between currents in opposite directions.

This field is uniform along wire 2 and perpendicular to it, and so the forceF2it exerts on wire 2 is given byF=IlBsinθF=IlBsinθwithsinθ=1sinθ=1:

F2=I2lB1F2=I2lB1.

By Newton’s third law, the forces on the wires are equal in magnitude, and so we just writeFfor the magnitude ofF2. (Note thatF1=−F2.) Since the wires are very long, it is convenient to think in terms ofF/l, the force per unit length. Substituting the expression forB1into the last equation and rearranging terms gives

Fl=μ0I1I22πr.Fl=μ0I1I22πr.

F/lis the force per unit length between two parallel currentsI1andI2separated by a distancer. The force is attractive if the currents are in the same direction and repulsive if they are in opposite directions. This force is responsible for thepinch effectin electric arcs and plasmas. The force exists whether the currents are in wires or not. In an electric arc, where currents are moving parallel to one another, there is an attraction that squeezes currents into a smaller tube. In large circuit breakers, like those used in neighborhood power distribution systems, the pinch effect can concentrate an arc between plates of a switch trying to break a large current, burn holes, and even ignite the equipment. Another example of the pinch effect is found in the solar plasma, where jets of ionized material, such as solar flares, are shaped by magnetic forces.

Theoperational definition of the ampereis based on the force between current-carrying wires. Note that for parallel wires separated by 1 meter with each carrying 1 ampere, the force per meter is

Fl=(4π×10−7T⋅m/A)(1A)2(2π)(1m)=2×10−7N/mFl=(4π×10−7T⋅m/A)(1A)2(2π)(1m)=2×10−7N/m.

Sinceμ0is exactly4π×10−7T⋅m/Aby definition, and because 1 T = 1 N/(A ⋅ m), the force per meter is exactly2×10−7N/m. This is the basis of the operational definition of the ampere.

THE AMPEREThe official definition of the ampere is: One ampere of current through each of two parallel conductors of infinite length, separated by one meter in empty space free of other magnetic fields, causes a force of exactly2×10−7N/mon each conductor.

Infinite-length straight wires are impractical and so, in practice, a current balance is constructed with coils of wire separated by a few centimeters. Force is measured to determine current. This also provides us with a method for measuring the coulomb. We measure the charge that flows for a current of one ampere in one second. That is, 1 C = 1 A ⋅ s. For both the ampere and the coulomb, the method of measuring force between conductors is the most accurate in practice.Section Summary

The force between two parallel currentsI1andI2, separated by a distancer, has a magnitude per unit length given by

Fl=μ0I1I22πr