# Magnetic Fields and Acceleration

1. Oct 8, 2006

### erik-the-red

Question

A particle with mass $$1.81 \times 10^{-3}\;{\rm kg}$$ and a charge of $$1.22 \times 10^{-8}\;{\rm C}$$ has, at a given instant, a velocity $$\vec{v} = (3.00 \times 10^{4}\;{\rm m}/{\rm s})\hat{j}$$.

What is the magnitude of the particle's acceleration produced by a uniform magnetic field $$\vec{B}=(1.63\;{\rm T})\hat{i} + (0.980\;{\rm T})\hat{j}$$?

I use the equation F = qv (CROSS) B.

I rewrite this as F = q*v*B*sin().

I plug in to get F = (1.22*10^(-8))*(3.00*10^(4))*(1.90) and get (6.96*10^(-4)).

I divide this by the mass, but my answer is wrong.

Obviously my procedure is not correct. What's up?

Last edited: Oct 8, 2006
2. Oct 8, 2006

### Päällikkö

You forgot the sine term.

3. Oct 8, 2006

### erik-the-red

So the angle is not 90 degrees?

4. Oct 8, 2006

### Päällikkö

Most certainly not. You have two vectors, one with only j component, but the other has i and j components. You could just compute the cross product, or you could figure out the angle. Or you could break the problem into two components: B parallel to v and B perpendicular to v.

5. Oct 9, 2006

### erik-the-red

I just figured it out. Thanks!