0
$\begingroup$

enter image description here So how can you determine the direction of the current here using just the right-hand rule but without using Lenz's law? The textbook says I should use the right-hand rule applied to the velocity of a charge but how does that help find the current? It only finds the resultant magnetic force on the moving charge. In this case, the magnetic flux is increasing, so applying Lenz's law means that the induced current's magnetic field will oppose the the flux which makes sense here with the direction of the current shown. But how can I just use the right-hand rule to find the current direction in this case? That's what my book says. Is there such a way?

$\endgroup$
0
$\begingroup$

The Lorentz force is given by $\mathbf{F} = q \mathbf{v}\times \mathbf{B}$ and indicates in which directions the charged particles (electrons here which have negative q) are pushed. If you consider an electron in the upper part of the loop, then we know the velocity direction of this electron and also the magnetic field at the location of the electron. If you then take the cross product of these two quantities, you know the force on the electron (via right-hand rule). The direction of the force will be parallel with the direction of the movement of the electron. The conventional current (movement of positive charges) will be in the direction opposite the flow direction of the electrons. This results in the current indicated on the figure.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ The cross product given above indicates the force on a positive charge and hence the direction of positive current flow. For any cross product, start with your right hand fingers in the direction of the first named vector (in this case,v). Hold your hand so that you can bend your fingers in the direction of the second named vector, (B). The direction of your thumb indicates the direction of the product (back along the top of the square loop). $\endgroup$ – R.W. Bird Mar 14 at 19:35

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.