The acceleration of a particle in uniform electro-magnetic field

A particle of charge $q$ and mass $m_0$ is subject to the action of a uniform electro-magnetic field $\vec{E}=(0,E,0), \vec{B}=(0,0,B)$ and at $t = 0$ moves with velocity $\vec{v}=(v_1,v_2,v_3)$.

What is the acceleration at $t = 0$ if $v_1 = v_3 = 0$?

-
What have you tried and where did you get stuck? Are you not able to formulate the equations of motion? Or not able to solve them? ...and if this is homework, please add a homework tag. – Marek Jun 22 '11 at 14:31
I'm not able to formulate the equation of accelation – Katy23 Jun 22 '11 at 14:35
Hint: The Lorentz force en.wikipedia.org/wiki/Lorentz_force – Qmechanic Jun 22 '11 at 14:40
Allow me to expand on Marek's comment. Do you understand how to deal with a single mass subject to two different forces? Do you understand the vector notation used here? Do you know how to find the force on a charge in an electric field? Do you know how to find the force on a charge moving in a magnetic field (and why I stuck the word "moving" in there)? Do you understand how to it all fits together? We understand that you haven't got an answer to the question---that is why you asked---but we want to know where you are stuck. – dmckee Jun 22 '11 at 15:06

As the comments said, this is a basic application of the Lorentz force. Written in your notation, we have the following.

$$\vec{F} = q[\vec{E} + (\vec{v} \times \vec{B})]$$

$$\vec{v} = \left( 0, v_2, 0 \right)$$ $$\vec{E} = \left( 0, E, 0 \right)$$ $$\vec{B} = \left( 0, 0, B \right)$$

The most likely thing you had trouble with would be the cross product I think. You can use a formula for this or your right hand. For the latter, in my own head, the magnetic field is "up" and the y-axis is toward my chest when I hold out my right hand. My middle finger points toward me, my index finger points up, and my thumb points to the right. So the answer is in the positive x-axis!

$$\vec{v} \times \vec{B} = \left( v_2 B , 0, 0 \right)$$

I will stop here because any more would certifiably be me doing it "for you". The idea of something being trivial is relative to the person, but everything that remains to answer your question should be trivial unless you are missing some fundamental part of physics instruction.

-