|
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 $v_1 = v_3 = 0$? |
|||||||||||||||
|
|
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. |
|||
|
|
