# Anybody know magnetic fields?

1. May 1, 2005

### Jon Wilson

i have a question that i need some help with:

at time t1, an electron is sent along the positive direction of an x axis, through both an electric field E and a magnetic field B, with E directed parallel to the y axis. Figure 28-34 gives the y component Fnet, y of the net force of the electron due to the two fields, as a function of the electron's speed v at time t1. the x and z components of the net force are zero at t1. Assuming Bx=0, find the magnitude E and B in unit-vector notion.

2. May 1, 2005

### OlderDan

Not enough information given for anyone else to do the problem without the figure. What have you done to try to solve this?

3. May 1, 2005

### Jon Wilson

there is a graph that has v as the x-axis and Fnet as the y-axis. it is a line that goes from x=0, y=-2 to x=50, y=0 to x=75, y=1. i know in ties into the formula v=E/B and possiably F=(qv)B. what i dont know is how to figure out what B is, if i knew how to do that i would be able to figure out the problem.

4. May 1, 2005

### Jon Wilson

the best i could come up with is E= -1.25x10^-38 N/C and B= (2.5x10^-2 T) ^k

5. May 1, 2005

### OlderDan

You are given that the fields are

$$\overrightarrow E = E_y \widehat j$$

$$\overrightarrow B = B_y \widehat j + B_z \widehat k$$

You know that the electric force is independent of velocity, so when you look at the graph where the velocity is zero, the net force must be from the electric field alone. You are also told that there is no net force in the x and z directions. What you must do is determine from the direction of the force, and the direction of the motion, which component of the magnetic field is producing the magnetic force. From the force versus velocity graph, you can determine the amount of force being contributed by the magnetic field for any velocity. You can pick one velocity (for example, the one that yields zero net force overall) and find the corresponding magnetic force (remember, the electric force is constant) and use that to determine the strength of the magnetic field.

6. May 1, 2005

### Jon Wilson

thanks for the explination. I have one more question for you:

A single-turn current loop, carrying a current of 4 A, is in the shape of a right triangle, with sides 50 cm, 120 cm, and 130 cm. The loop is in a uniform magnetic field of magnitude 75 mT whose direction is parallel to the current in the 130 cm side of the loop. What is the magnitude of the magnetic force on (a)the 130 cm side, (b) the 50 cm side, and (c) the 120 cm side? (d) What is the magnitude of the net force on the loop?

I worked it out and the answers I got were: (a) Fb= 0 (b) .0138 N (c) .0138 N (d) 0. Does this look correct to you?

7. May 3, 2005

### OlderDan

a) is correct. b) and c) should not be the same, so d) cannot be right. How are you doing b) and c)?