Question

How do you find the magnitude and direction of the electric field and magnetic field?

When at rest a pattern experiences a net electromagnetic force of 8.5x10^-13 N pointing in the x-direction. When the proton moves with a speed of 1.5x10^6 m/s in the positive y direction the net electromagnetic force decreases to 7.5x10^-13 N, still pointing in the positive x direction.

Answer 1

Have a look if it's understandable (AND if it seems correct!).

Explanation:

Ok, this is complicated...!

I would try using Lorentz Law:

`vecF=qvecE+qvecvxxvecB`

with the vector product:

I think in both cases you have a proton (with charge `q=1.6xx10^-19C`)!

1) at rest: `vecv=0`, the proton (its charge) is experiencing the `E` field around it and, as a consequence, the force:
`vecF=qvecE`
(`vecB=0`)
using moduli (say, along the `x` axis)
`E=F/q=(8.5xx10^-13)/(1.6xx10^-19)=5.3xx10^6V/m`
along the `x` direction and zero magnetic field.

2) Use the same approach with the given value of velocity remembering that now you'll have a magnetic field `vecB` perpendicular (due to the vector product) to the force (in the `x` direction) and velocity (in the `y` direction) and so in the `z` direction.

We need to consider the decrease in force so we need a negative value in the product: `qvecvxxvecB` (probably the `vecB` field will be in the negative `z` direction giving a negative modulus for the vector product).
We can write using our data into: `vecF=qvecE+qvecvxxvecB` considering the moduli:

`7.5xx10^-13=8.5xx10^-13+1.6xx10^-19*1.5xx10^6*B`
rearranging:
`B=-0.416~~-0.42T`

Here it is a possible visual representation: