Let's suppose there's a horizontal and uniform magnetic field $B$, and we introduce from below the field a square coil of mass $m$, side length $a$ and resistance $R$ in it with a inicial velocity $v_0$. I'm asked to find the time it takes for the coil to stop its movement.

I'll try to explain what I tried to do. Since the coil is going up entering the field, its magnetic flux is changing, therefore appearing an inducted current in the coil. This current $I$ and the magnetic field $B$ will cause a magnetic force which will only affect the upper part of coil, since the forces on the sides will cancel and the lower part hasn't entered the field yet. Since the magnetic force $F_m$ drags the coil down, we'll hace the equation:

$$ mg+F_m= m\dfrac{dv}{dt} $$

Where $F_m=ILB$ by Laplace's law and $I=\dfrac{\varepsilon}{R}$.

Now, by Faraday's law, $\varepsilon=-\dfrac{d\Phi}{dt}$; we can say that $\Phi=\int BdS$ (with the dot product already done). And here comes my problem, because $S=L\cdot y$, where $y$ is the "height" of the coil already in the field, so $dS=Ldy$, but I can´t find a way to deal with that $dy$, since I can't assume that it has constant velocity nor acceleration, I can't seem to find a way to correlate it with the time passed.

I tried it by assuming that we can say $dy=d(vt)=tdv+vdt$, so our integral would be:

$$ \int BdS = \int B\cdot a\cdot dy = \int B \cdot a \cdot d(vt) = \int _{v_0} ^{0} B \cdot t \cdot dv + \int _0 ^{t_f} B \cdot a\cdot v\cdot dt $$

Because we're integrating from the point of $t=0, v=v_0$ to a certain $t_f$ and the velocity at that moment.

Solving that integral would give us $\Phi=-B\cdot a\cdot t\cdot v_0+B\cdot a\cdot t_f\cdot v$, and deriving it:

$$ \varepsilon = \dfrac{d}{dt}\left( B\cdot a\cdot v_0\cdot t-B\cdot a\cdot v\cdot t_f\right) = B \cdot a\cdot v_0 - B\cdot a\cdot t_f\dfrac{dv}{dt} $$ And we obtain $I=\dfrac{B \cdot a\cdot v_0}{R}-\dfrac{B\cdot a\cdot t_f}{R}\dfrac{dv}{dt}$ and $F_m=IaB=\dfrac{B^2 \cdot a^2\cdot v_0}{R}-\dfrac{B^2\cdot a^2\cdot t_f}{R}\dfrac{dv}{dt}$

Now, getting back $$ mg -F_m = mg-\dfrac{B^2 \cdot a^2\cdot v_0}{R}+\dfrac{B^2\cdot a^2\cdot t_f}{R}\dfrac{dv}{dt} = m\cdot \dfrac{dv}{dt} $$ And there it would be just grouping terms and integrating, and the final result I got was $t_f=\dfrac{v_0}{g}$.

I don't think it is correct because I'm not sure the assumptions I made about that $dy$ can be made, but I have no other idea. I'm pretty sure there must be an easier way, because this is a first year of physics degree problem so it can't require a high level of maths.

Thank you so much in advance, and sorry about the messy formatting, it was my first question here .


closed as off-topic by G. Smith, John Rennie, Kyle Kanos, GiorgioP, Aaron Stevens Jul 11 at 14:25

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