# Combination of electric and magnetic fields

1. Feb 28, 2005

### bemigh

Hey, i have this question to solve..
An electron has a velocity of 1.62×104 m/s (in the positive x direction) and an acceleration of 2.25×1012 m/s2 (in the positive z direction) in uniform electric and magnetic feilds. If the electric field has a magnitude of 19.1 N/C (in the positive z direction), what is the y component of the magnetic field in the region?

Ok, so..
The electric field is in the direction of the acceleration, so i solved for the theoretical acceleration caused by the electric field, which turned out to be 3.35x10^12 m/s^1. This is MORE than the actual acceleration, therefore, the magnetic force must be causing the electron to accelerate 1.10x10^12 m/s^2 in the negative z direction. Using this acceleration and mass of the electron, the magnetic force is 1.00625e-18, and using F=qv X B, i found the magnetic field in the Y direction to be 3.88x10^-4 T, which is ultimately wrong...
where did i go wrong?
Cheers
Brent

2. Feb 28, 2005

### Gamma

Remember, that 2.25 X 10^12 is the net accelaration due to electric and magnetic feild. Accelaration due to the electric field is in the -z direction and is = 3.35x10^12 m/s^1. So accelaration caused by the magnetic field should be (2.25 X 10^12 +3.35x10^12) m/s^1 in the +z direction. If you use the lorentz force equation, there is no chance of making a mistake in directions.

Start from the lorentz Equation.

$$F= ma \vec z = -eE\vec z -e (vXB)$$

$$ev (\vec x X B) = (-eE - ma)\vec z$$

$$B_y = \frac{-eE - ma}{ev}$$

3. Feb 28, 2005

### bemigh

still problems

Hey,
using the larentz equation, i got an answer of 7.89 x 10^-4 T, which is still wrong... any ideas?
Cheers

4. Mar 1, 2005

### Gamma

How did you come up with this number? Could you post the answer?

Substituting in the formula that I got earlier, By= -1.97 X 10^-3 T.

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