Will it ever stop?

Electricity and Magnetism Level 3

A vertical conducting rod of mass $m$ and length $L$ is free to slide on two smooth horizontal conducting rails as shown, where there is a magnetic field of intensity $B$ directed inside the screen and the resistance of the resistor is $R$ . The rod is given an initial velocity of $v_0$ . Find the total distance travelled by the rod.

Assume that no other forces other than the magnetic force are acting on the rod.

Inspiration and credits for the image .

\dfrac{mRv_0}{B^2L^2}e^{-\frac{B^2L^2}{mR}}

\dfrac{2mRv_0}{3B^2L^2}

None of these

\dfrac{mRv_0}{B^2L^2}

\infty

\dfrac{B^2L^2}{mRv_0}

\dfrac{mRv_0}{B^2L^2}\left(1-e^{-\frac{B^2L^2}{mR}}\right)

1 solution

Alan Enrique Ontiveros Salazar
Jul 1, 2016

Let $v$ be the velocity at any point of the movement of the rod. We see that the induced emf is $\epsilon=BLv$ . Then, the current is $I=\dfrac{\epsilon}{R}=\dfrac{BLv}{R}$ and the magnetic force, which is directed to the left, has a magnitude of $F_M=ILB=\dfrac{B^2L^2v}{R}$ .

Now, if we apply Netwon's second law we get: $-F_M=ma$

$-\dfrac{B^2L^2v}{R}=m\dfrac{\mathrm dv}{\mathrm dt} \implies -\dfrac{B^2L^2}{mR}\mathrm dt=\dfrac{\mathrm dv}{v}$

Integrate both sides:

$\displaystyle -\dfrac{B^2L^2}{mR} \int_{0}^{t}\mathrm dt=\int_{v_0}^{v}\dfrac{\mathrm dv}{v} \implies -\dfrac{B^2L^2t}{mR}=\ln\left(\dfrac{v}{v_0}\right)$

Solve for $v$ :

$v=v_0e^{-\frac{B^2L^2t}{mR}}$

Find the displacement in terms of time:

$\displaystyle r=\int_{0}^{t}v\;\mathrm dt=\dfrac{mRv_0}{B^2L^2}\left(1-e^{-\frac{B^2L^2t}{mR}}\right)$

Finally, we see that when $t\rightarrow \infty$ the displacement converges to:

$\displaystyle \lim_{t \rightarrow \infty} r=\boxed{\dfrac{mRv_0}{B^2L^2}}$

Will it ever stop?

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