# Moving conducting bar in (changing) magnetic field

In my exam today I've been given this problem, yet even with the results at hand I simply can't warp my head around it;

Given the picture below, a bar is placed on two conducting rails with a resistor (in the form of a single thin wire) at the left. Now at t=0 the magnetic field changes which causes the bar to move. (and a current flows through the "circuit").

The current can easily be found by Lenz's law:

$$\mathscr{E} = -\frac{\mathrm{d}\Phi_b}{\mathrm{d}t} = -\ell\frac{\mathrm{d}\left( Bx \right )}{\mathrm{d}t} = -\ell x\frac{\mathrm{d} B }{\mathrm{d}t} -\ell B v$$ Now this is follow up text for the problem

It may be assumed that the conducting bar has a negligible mass and friction, so it will immediately move to a position $x$ in which there is no force exerted on it.

b) Explain why the relation between the actual position $x$ and the actual value of the magnetic field $B$ is given by $Bx = B_0x_0$.

What did the prof try to explain here? No force means constant speed, so not a "position" at all right?
Well the result sheet shows:

No force means $\frac{\mathrm{d}(Bx)}{\mathrm{d}t} = 0$ Hence $Bx$ is constant .....

what???? Sorry but maybe it stems from the fact that I already didn't understand what "no force" meant in the text but I can't understand this.

(my original answer was wrong - I misunderstood the assignment.)

What the problem was trying to demonstrate is that in quasi-static (slow) changes of field around circuits with negligible resistance, magnetic flux through the circuit remains approximately the same. The rod will experience weak force due to induced electric field and will move in such a way as to preserve the magnetic flux through the circuit.

Why is the magnetic flux constant?

As the electric field induces weak current in the rod, magnetic force will pull the rod in the indicated direction with force $IB\mathscr{l}$. The rod will accelerate in the first instants, but since it is light-weight, it will quickly reach velocity that is sufficient for the magnetic electromotive force (opposes the current) in the $rod$ to counter-act the electric electromotive force of the changing external field in the $circuit$.

The induced electric electromotive force can be found from the integral version of the Faraday law and is $$U_{emf,el.} = \mathscr{l}x\dot{B}.$$ The magnetic electromotive force acts only when the rod moves and is known to be equal to $$U_{emf,mag.} = \mathscr{l}vB.$$

When the equilibrium is attained, the two emf's balance each other: $$U_{emf,el.} + U_{emf,mag.} = 0,$$ and since $v = \frac{dx}{dt}$, $$\frac{d}{dt}\left( xB\mathscr{l}\right) = 0,$$ which means that the magnetic flux through the loop (bound to the metallic circuit and moving with it) remains constant.

As it has been said, the force is given by the Laplace's law $$\vec F=\int \vec I\wedge \vec B\,\mathrm dx=I\ell B\,\vec\imath$$ where $\vec\imath$ is the horizontal unit vector from left to right on the figure. Now, Lenz's law give a difference of potential (a voltage) of $E=-\ell\,{\mathrm d(Bx)}/{\mathrm d t}$ so the current is $$I=\frac ER=-\frac{\ell}{R}\frac{\mathrm d(Bx)}{\mathrm d t}$$ hence the force $$F=\|\vec F\|=\frac{\ell^2}{R}B\frac{\mathrm d(Bx)}{\mathrm d t}$$ is zero if $Bx$ is constant.

The force acting on the conducting bar will be given by:

$$F=-ilB$$

But it is given that the bar has a negligible mass.So there can't be any force acting on the bar otherwise it would experience an "infinite" acceleration.

This in turn implies that the current flowing through the bar must always be zero.So there can't be a changing EMF and hence the time derivative of the flux will be zero.Thus,we have $$\frac{d\Phi}{dt}=0$$ $$Bx=\text{constant}$$.

So in summary, zero mass of the bar$\implies i=0 \implies \frac{d\Phi}{dt}=0 \implies Bx=Bx_0$