0
$\begingroup$

The picture shows the situation described

I have the above question and I have though about it every way and can't seem to find out. So what I have concluded (ignore the answers on the img) Is that the force is on the z axis because the cross product of the two would be the normal of the plane it lies on which is into or outof the screen.

Thus I am saying that first one is false because it is on the z axis, the finite amount I said true because it leaves the magnetic field when being pushed perpendicularly away. The next one because it IS normal to the velocity. Lastly the y-component is unchanged again because the force points the z direction not the y.

Any reasons to what I am thinking wrong any help would be appreciated.

$\endgroup$
  • $\begingroup$ Just because a force is always normal to the direction of travel doesn't mean that the object will necessarily travel in a circular orbit. $\endgroup$ – DumpsterDoofus Mar 14 '14 at 20:54
  • $\begingroup$ @DumpsterDoofus Okay that makes sense actually I was thinking perhaps the circular part would cause me issues I will try it out $\endgroup$ – capa_matrix Mar 14 '14 at 21:01
  • $\begingroup$ @DumpsterDoofus So according to the rest of my logic using false,true,false,true is still wrong so are there any other flaws I can't see $\endgroup$ – capa_matrix Mar 14 '14 at 21:03
1
$\begingroup$

You can simply decompose the velocity in its $v_x$ and $v_y$ components. The first one will remain unaffected (do a Galileo transformation, if you are not sure, and observe from a frame moving at $v_x(0)$ in the $x+$ direction).

The perpendicular component to the force will have a constant modulus, but its direction will change uniformly, so the particle will describe a helicoidal trajectory.

No one of the options is actually correct, but the last one is the closest (could just be bad wording). For the others:

  • The force is in the $YZ$ plane, never in $\vec x$
  • Magnetic forces cannot do work.
  • The particle has a constant $v_x$, so the trajectory cannot be circular.
$\endgroup$
  • $\begingroup$ Thankyou for the help you were right in that all of them were false. Thanks again. $\endgroup$ – capa_matrix Mar 14 '14 at 21:24

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.