0
$\begingroup$

I have this problem where i should find the direction and magnitude of the electric and the magnetic force on the electron. And then I'm supposed to find the direction and magnitude of the acceleration.
E = 1000 N/m B = 2,5 T v = 500 m/s

X |X X| X
 ¤->
X |X X| X
X |X X| X
  V   V

I've found the electric force by Fe = E*q. With direction out of the screen by the right hand rule. Magnetic force = qvb in the y direction (meaning up) by the right hand rule. Now Im supposed to find the acceleration but I'm a bit stuck here. What i'm struggling with is to combine the two forces on the electron. Can someone get me in the right direction?

$\endgroup$
  • $\begingroup$ Am I missing something? Don't you just add the two forces (vector addition) to get a total force $\mathbf F$ then use $\mathbf F = m \mathbf a$? $\endgroup$ – John Rennie Apr 27 '16 at 16:18
  • $\begingroup$ So F = Fm + Fe? $\endgroup$ – KimR Apr 27 '16 at 16:20
  • $\begingroup$ Using vector addition, yes. It isn't obvious from your diagram what the direction of $\mathbf E$ is. $\endgroup$ – John Rennie Apr 27 '16 at 16:22
  • $\begingroup$ it's down. The answer made sense. Guess i was lost in my head a moment there. Thanks for the simple but valuable input :) $\endgroup$ – KimR Apr 27 '16 at 16:23
1
$\begingroup$

You just use vector addition and Newton's Second Law. For example, if you have $$\overrightarrow{F}_E=qE\hat{x},\space\space\space\overrightarrow{F}_B=qvB\hat{y}$$ then your total force $F_{tot}=F_E+F_B$ is just $$\overrightarrow{F}_{tot}=qE\hat{x}+qvB\hat{y}$$ Since $\hat{x}$ and $\hat{y}$ are totally linearly independent, these terms cannot be combined. Then using $F=ma$ you can deduce that $$\overrightarrow{a}=\frac{q}{m}(E\hat{x}+vB\hat{y})$$ but if you wanted the magnitude of the acceleration, you just take $$a=\sqrt{\overrightarrow{a}\cdot\overrightarrow{a}}=\frac{q}{m}\sqrt{E^2+v^2B^2}$$

$\endgroup$
  • $\begingroup$ I'm not sure what you mean. The force exerted by an electric field is just $\overrightarrow{F}_E=q\overrightarrow{E}$, so if you know the field finding the force is trivial. $\endgroup$ – ocket8888 May 3 '16 at 7:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.