0
$\begingroup$

While deriving the formula for motional emf, we consider an equilibrium point where the force on the electron due to the magnetic field (Lorentz force) is equal to the force of electric field.

Hence $f_m$=$f_e$

ie. evB•sin 90 = eE (e is electronic charge)

this would give vB= E

Isn't electric field equal to - $\Delta$ $\frac Vl$?Wouldn't that result in $\epsilon$ to be negative (ie. - Bvl), which is obviously not the case? What am I missing here?

Improve this question
$\endgroup$
2
  • $\begingroup$ What you should do at the start is define some axes. $\endgroup$
    – Farcher
    Commented Aug 17 at 16:07
  • $\begingroup$ @Farcher Sorry for being vague. But I'm considering a conducting rod (length l) moving with some velocity v where magnetic field B is perpendicular and goes into the plane. So force f(e) is along positive j cap, and force f(m) is along negative j cap (if that's what you meant). $\endgroup$
    – Mel
    Commented Aug 17 at 16:12

3 Answers 3

Reset to default
0
$\begingroup$

So if rod moves from left to right, electrons move down(right hand rule but negative charge) more electron down, potential difference. BUT, isn't magnitude of electric force equal to magnitude of magnetic field, this then removes the negative sign in -delts V/L. Electric force acts upwards, magnetic force acts downwards, this direction does not come into effect in the equation F(magnetic) = F(electric). This means direction of field is not important(-/+)

Improve this answer
$\endgroup$
6
  • 1
    $\begingroup$ Wait, here charge is negative that means even if we go by your formula, E would by -(-E) making it +E $\endgroup$
    – Curio
    Commented Aug 17 at 16:43
  • $\begingroup$ The sign of most of the terms are determined by using a unit positive test charge, so since we are talking about electrons here, i guess that means minus minus $\endgroup$
    – Curio
    Commented Aug 17 at 16:49
  • $\begingroup$ how could you take positive charge here, i mean aren'yt we talking about moving electrons ? $\endgroup$
    – Curio
    Commented Aug 17 at 16:57
  • $\begingroup$ Ah, right, got a fundamental thing wrong. Thanks for the help. $\endgroup$
    – Mel
    Commented Aug 17 at 16:59
  • $\begingroup$ E = -Delta V/L is general formula. Field is force per charge(positive) when we talk about negative charges won't the equation change in sign ? this is what i think, thanks. $\endgroup$
    – Curio
    Commented Aug 17 at 17:00
0
$\begingroup$

Let the charge on the charge carrier be $q$ which can either be a positive or negative quantity.

enter image description here

$\vec F_{\rm m} = q\,\vec v\times \vec B = q\,v\,\hat i \times B\,\hat k = -q\,v\, b\,\hat j$ and $\vec F_{\rm e} = q \,\vec E = q\,E\,\hat j$.

$\vec F_{\rm m}+\vec F_{\rm e}=0 \Rightarrow v= \frac EB$

$\vec E = E\,\hat j = - \frac{\Delta V}{\Delta y}\,\hat j = -\frac{V_{\rm B}- V_{\rm A}}{\ell}= \frac{V_{\rm A}- V_{\rm B}}{\ell}$ and note that $V_{\rm A}> V_{\rm B}$ so the component of the electric field $E$ is positive.

Improve this answer
$\endgroup$
0
$\begingroup$

I assume that you have in mind an electric field arising from displacement of charge carriers in a conductor due to the magnetic force on them as the conductor is moved.

Equilibrium is reached when the electric force on a charge carrier is equal and opposite to the magnetic force. The potential rises as we go along the conductor in the opposite direction to that of the electric field ($\frac{\partial V}{\partial y}=-\mathbf E_y$), that is in the same direction as the magnetic force per unit positive charge ($\mathbf v \times \mathbf B$). In other words the end of the conductor to which $\mathbf v \times \mathbf B$ points is at a more positive potential than the other end. That's as it should be.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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