Problem :

Two wires run parallel to each other, each with a current of 10⁹ esu/sec. If each wire is 100 cm long, and the two wires are separated by a distance of 1 cm, what is the force between the wires?

Solution for Problem 1 >>

This is the simplest case of magnetic interaction between currents, and we simply plug in values to our equation:

F = = = .222 dynes

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Problem :

Three wires, each with a current of i , run parallel and go through three corners of a square with sides of length d , as shown below. What is the magnitude and direction of the magnetic field at the other corner?

Three wires run a current into the page. What is the magnetic field at point P ?

Solution for Problem 2 >>

To find the net magnetic field, we must simply find the vector sum of the contributions of each wire. The wires on the corners contribute a magnetic field of the same magnitude but are perpendicular to each other. The magnitude of each is:

B =

The wire diagonal from point P has a magnetic field at P of magnitude:

B =

All these fields point in different directions, and to find the total field we must find the vector sum of each field. Shown below are the directions of the contributions of each wire:

The fields from each wire at point P

The directions of these fields are found by using the second right hand rule. >From our diagram, we sum the x and y components of each field:

B x	=	- B 2 - B 3sin 45 o = - - = -
B y	=	- B 1 - b 3sin 45 o = - - = -

Notice from the symmetry of the problem that the x and y components have the same magnitude, as expected. Also from symmetry we can tell that the net force will act in the same direction as the field from B ₃ , down and to the left. Its magnitude comes from the vector sum of the two components:

B _T = = 3

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Problem :

Compass needles are placed at four points surrounding a current carrying wire, as shown below. In what direction does each needle point?

Compass needles are placed at four points in a plane perpendicular to the wire.

Solution for Problem 3 >>

Compasses in the presence of a magnetic field will always point in the direction of the field lines. Using the right hand rule we see that the field lines flow counterclockwise, as seen from above. Thus the compasses will point as such:

The direction of the four compasses placed around a current-carrying wire

Compasses are often used to find the direction of a magnetic field in a given situation.

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Problem :

What is the force felt by a particle with charge q travelling parallel to a wire with current I , if they are separated by a distance r ?

Solution for Problem 4 >>

We have derived the force felt by another wire, but have not derived it for a single particle. Clearly the force will be attractive, as the single charge can be seen as a "mini current" running parallel to the wire. We know that B = , and that F = , since the field and the velocity of the particle are perpendicular. Thus we simply plug in our expression for B :

F =

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Problem :

Two parallel wires, both with a current I and length l , are separated by a distance r . A spring with constant k is attached to one of the wires, as shown below. The strength of the magnetic field can be measured by the distance the spring is stretched due to the attraction between the two wires. Assuming that the displacement is small enough that at any time the distance between the two wires can be approximated by r , generate an expression for the displacement of the wire attached to the spring in terms of I , r , l and k .

Two wires, one connected to a spring, shown before the spring is stretched to the force between the wires.

Solution for Problem 5 >>

The spring will reach its maximum displacement when the force exerted by one wire on the other is in equilibrium with the restoring force of the spring. At its maximum displacement, x , the distance between the two wires is approximated by r . Thus the force on one wire by the other at this point is given by:

F =

Recall also that the restoring force of the spring is given by

F = kx

The wire is in equilibrium when these two forces are equal, so to solve for x we relate the two equations:

	=	kx
x	=

Although we did use an approximation to find the answer, this method is a useful way of determining the strength of the magnetic force between two wires.

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