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
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Solution: Click For PDF Version It is given that four long copper wires are parallel to each other and are
arranged in a square. They carry equal currents
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
The magnetic field at C due to B will be According to the Lorentz force law, the force on a segment
As the current in C is in the direction
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
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
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