Problem #0199 Electricity & Magnetism Sub-menu Problem #0201 Chapters Chapters

470.

Problem 35.35 (RHK)

Four long copper wires are parallel to each other and arranged in a square, as shown in the figure. They carry equal currents out of the page, as shown. We have to calculate the force per meter on any one wire. We have to give its magnitude and direction. We may assume that and . (In the case of parallel motion of charged particles in a plasma, this is known as the pinch effect.)

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It is given that four long copper wires are parallel to each other and are arranged in a square. They carry equal currents out of the page. We have to calculate the force per unit meter on any one of the wires. Current in the wires,

,

and the separation between the wires,

.

As the wires are parallel and current is flowing in the same direction in each one of them, the force between any pair will be attractive and will act along the perpendicular line joining the wires. In the figure we have labelled the four wires as A, B, C and D. Let us compute the force per unit length on the wire C.

We will derive the formula for the force between two current carrying parallel wires by computing the force on C due to B.

We fix the coordinate system by drawing unit vectors and , as shown in the figure.

The magnetic field at C due to B will be

According to the Lorentz force law, the force on a segment carrying current in a magnetic field is given by

As the current in C is in the direction , the force per unit length on C due to B will be

Similarly, the force per unit length on C due to D will be

The magnetic field at C due to A is

Therefore, force per unit length on C due to A will be

Therefore, the force per unit length on wire C due to wires A, B and D will be

is a unit vector pointing toward the centre of the square from C.

We now put in the data and calculate the magnitude of the force per unit on any one of the wires because of the other three.

The direction is toward the centre of the square. From symmetry, we conclude that force per unit length on any wire due to the other three will be of and will be pointing toward the centre of the square formed by the four current carrying wires.