0
$\begingroup$

We say that Force exerted by a current carrying wire on charge $ q $ moving with velocity $v$ is: $F=qvB\tag*{}$ (Where $B$ is the magnetic field , and electric Force is 0 because wire is neutral.)

But an observer moving in the direction of charge, with velocity $v$, would see the charge at rest. So, in this frame, no Force should act on the charge?

Please tell me where am I wrong?

$\endgroup$
1
$\begingroup$

If charge $q$ in a system $S$ moves with velocity $v$ in a direction perpendicular to a magnetic field $B$ (and there is no electric field in $S$), it experiences in $S$ a Lorentz force $$F=qvB$$ If you are in an inertial system $S'$ moving with the charge at constant velocity $v$ then you see an electrical force on the charge $$F'=qvB'$$ where $$B'=\frac{B}{\sqrt{1-v^2/c^2}}$$ is the magnetic field in $S'$ in the same direction as in $S$. For small velocities $v<<c$, the magnetic field in $S'$ is approximately $B'=B$. This means that, in this approximation, the Lorentz force $F$ on $q$ in $S$ is equal to the electrical force $F'$ on $q$ in $S'$ $$F=F'$$ Thus what is a magnetic force on the moving charge in $S$ appears as a purely electrical force in $S'$. This follows from the Lorentz transformation of electric and magnetic fields in inertial systems moving with velocity $v$ relative to each other. See, e.g., Chapter 12.3 Relativistic Electrodynamics, in D.J. Griffiths, Introduction to Electrodynamics, 3rd edition, 1999.

$\endgroup$
  • $\begingroup$ If you change to $S'$, why do you use $v$, not $v'$ in the expression for force? In the question exactly this point is not understood. $\endgroup$ – queezz Mar 2 '18 at 5:18
  • $\begingroup$ @queezz - Because it was assumed that the charge doesn't move in $S'$. See the text above.. $\endgroup$ – freecharly Mar 2 '18 at 17:32
  • $\begingroup$ I just think it could be much more obvious if you explicitly state the transformation of the electromagnetic forces. $\endgroup$ – queezz Mar 2 '18 at 19:18
  • 1
    $\begingroup$ @queezz - You are right, I just wanted to keep the answer short. $\endgroup$ – freecharly Mar 2 '18 at 20:27
0
$\begingroup$

Expanding the freecharlys answer.

When changing from one inertial frame to another don't forget that electric and magnetic fields are two manifestation of one electromagnetic force.

Let us have two frames, $S$ where charge is moving and $S'$ where it is at rest. Then for $S'$ we can wright down $$ \mathbf{E_\parallel}' = \mathbf{E_\parallel} \\ \mathbf{B_\parallel}' = \mathbf{B_\parallel} \\ \mathbf{E_\bot}' = \gamma \left( \mathbf{E}_\bot + \mathbf{v} \times \mathbf{B} \right) \\ \mathbf{B_\bot}' = \gamma \left( \mathbf{B}_\bot - \frac{1}{c^2} \mathbf{v} \times \mathbf{E} \right) $$ where $\gamma \ \overset{\underset{\mathrm{def}}{}}{=} \ \frac{1}{\sqrt{1 - v^2/c^2}}$, see Wiki for more details.

Important part now is $\mathbf{E_\bot}' = \gamma \left( \mathbf{E}_\bot + \mathbf{v} \times \mathbf{B} \right)$, even if $\mathbf{E} = 0$ in the frame $S$, in the frame $S'$ electric field would not be zero. Assuming velocity of the charge $v<<c$, in your case $\mathbf{E}' = \mathbf{v}\times \mathbf{B}$.

So if $\mathbf{v} \perp \mathbf{B}$, the force in $S'$ will be defined as follows: $$F' = qE = qvB = F $$

Force would not change, though the electromagnetic field in frame $S'$ will. And it will have an electric component now, which will act on the charge at rest.

$\endgroup$
-1
$\begingroup$

Quote from undergraduate physics textbook from Knight: Basically, when you trying to observe in a "moving" reference frame, $E$ and $B$ field started to rotate by an imaginary angle. In this case, $B$ field became $E$ field, thus in stead of the magnetic force, there ought to be a equivalent electronic statics force.

$\endgroup$

Your Answer

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