1
$\begingroup$

enter image description here

Given a point charge $q$ near a wire carrying current as shown, what would be the effect of the magnetic field produced by the wire on this point charge $q$? I think the sign on $q$ is important so let's consider both cases in which the charge $q$ is positive and negative separately.

$\endgroup$
1
$\begingroup$

Use the right-hand rule: the thumb goes upwards with the conventional current of the wire. This means the magnetic field will be going counter-clockwise when viewed from the top.

At the location of the charge, the magnetic field points out of the screen. Now, use Fleming's left-hand rule: The thumb will be the direction of the force which we want, the index finger is the magnetic field (out of the screen), and the middle finger is the movement of the charge, however, the charge is not moving! So no force!

Mathematically, this is seen as: $\vec{F}=q\vec{v}\times\vec{B}$. Since the charge is not moving, $\vec{v}=0$, and thus $\vec{F}=0$. Since no force is acting on the stationary charge there will be no acceleration and no movement.

Note: the only difference between a positive and negative charge in relation to this would be the direction of the charge's movement in the left-hand rule. If it is positive, then it is the normal direction. If negative, then it is opposite to the direction. This is because of the definition of conventional current used in such rules.

$\endgroup$
  • $\begingroup$ Makes sense, so based on this, let's say q is moving towards the wire, the force on it will then be upwards right? And if it is an electron bearing negative charge, the force will be downward? $\endgroup$ – Arkilo Jul 16 at 19:52
  • $\begingroup$ @Arkilo Sorry I made a typo about which finger represents what for the left hand rule. The middle finger is movement and index is magnetic field. So the result would be the other way round for the charge signs. $\endgroup$ – Alaz Jul 16 at 19:59
  • 1
    $\begingroup$ @Arkilo another interesting thing to note is that if the charge moved parallel to the wire it would rotate around the wire since force points inwards, leading to uniform circular motion. Also if you're happy, please accept the answer. $\endgroup$ – Alaz Jul 16 at 20:00
  • 1
    $\begingroup$ Interesting, so simple based on intuition, in the first case, the magnetic fields will be interfering destructively between the wires and have weaker overall field as compared to the sides. This will cause the wires to be attracted to each other and in case of opposite flow, the magnetic field between then will be stronger as compared to the sides so they will be repulsed from each other. Did I get it right? $\endgroup$ – Arkilo Jul 16 at 20:20
  • 1
    $\begingroup$ @Arkilo Ahhh good ol' physics. The best way to solve a physics problem is always to simplify it down to what you know and keep it simple. $\endgroup$ – Alaz Jul 16 at 20:55

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.