# Electric Field near current carrying wires

1. Dec 3, 2007

### Reshma

1. The problem statement, all variables and given/known data
Two parallel infinitely long wires separated by a distance D carry steady currents I1 and I2 (I1>I2) flowing in the same direction. A positive point charge is moves between the wires parallel to the currents with a speed v at the distance D/2 from either wire. What is magnitude of an electric field that must be turned on to maintain the trajectory of the particle is proportional to?

2. Relevant equations
Magnetic attraction force per unit length $$f_m = {\mu_0 \over 4\pi} {I_1 I_2 \over D}$$
If $\lambda$ is the linear charge density the electric force of repulsion per unit length is: $$f_e = \frac{1}{2\pi \epsilon_0}{\lambda \over D}$$

3. The attempt at a solution
Without the charge moving the force between the wires will be attractive. The moving charge with exert a repulsive force. How do I bring the electric field in this picture? Use Gauss's law?

Last edited: Dec 3, 2007
2. Dec 3, 2007

### Gokul43201

Staff Emeritus
I think there's a mistake in this problem. For starters, the last line of the question is hard to interpret exactly and besides that, the answer is the trivial result.

But in any case, you do not need to worry about the force between the wires. You are only asked to ensure that the moving charged particle keeps moving in a straight line. What is the net force on the charged particle due to the currents in the wires? How will you calculate this?

3. Dec 4, 2007

### Reshma

The force on the wires will be:
$$\vec F_1 = I_1\vec L \times \vec B$$
$$\vec F_2 = I_2\vec L \times \vec B$$
Since the charge is moving the force on the charged particle will be $$\vec F = q\vec v \times \vec B + q\vec E$$
How are the forces balanced?

Last edited: Dec 5, 2007
4. Dec 4, 2007

### Gokul43201

Staff Emeritus
We care only about the force on the particle. The Lorentz force depends on the angle between the velocity of the charge and ...what? What is the magnitude of the B-field exactly midway between the wires?

5. Dec 5, 2007

### Reshma

Lorentz force depends on the angle between the velocity and magnetic field of the charged particle. The magnitude of the magnetic field at D/2 will be:
$$B = {{\mu_0 (I_1 + I_2)} \over {\pi D}}$$
I have taken I1 + I2 due to the attractive nature of the magnetic force. Should I put this value for B in the Lorentz force equation? Also the magnitude of the magnetic force will be maximum since the particle is moving perpendicular to the field.

Last edited: Dec 5, 2007
6. Dec 5, 2007

### Gokul43201

Staff Emeritus
This is not correct.

Use the right-hand rule to figure out the direction of the magnetic field due to the current in each wire separately. What do you notice?

7. Feb 15, 2008

### neelakash

It is an old question...and it has not been solved propely.So,I am suggesting a method.May be that it is not a method that other people would appreciate.But this works well in exam hall.

[By the way, reshma missed something in the last part of the question, for which this looked rather diffcult.The question is an mcq and the options are:

(a) (I1-I2)v/D (b) (I1+I2)v/D (c) (I1-I2)v²/D² (d) (I1+I2)v²/D²

And Gokul43201 will find his answer.]

First start with dimensional analysis.(a) and (b) has the same dimension C/s² and (c) and (d) has the same dimension C/s³

Clearly (c) and (d) cannot be proportional to the electric field, the constant of proportionality being (1/(4pi*epsilon)) type.The constant of proportionaliy must be indeependent of time.Note that the question specifies STEADY current and we must deal with time independent electric field to satisfy the conditions imposed by the question.

Then we are left with (a) and (b).Since,I1 > I2,the magnetic force must be different on the charged particle and the particle should bend sideways.The electric field that CAN make the charge move along the straight line must be proportional to |I1-I2|,thus preventing the deflection of the charged particle due to non-uniform magnetic force.

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