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\begin{document}
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\heading{Induction}
\auname{I.D. 034660811}
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{\bf The problem:}
\Dn
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A rod in lenght $l$ is sliding with a constant velocity $v$ on two
conducting rails. The system is in inhomogeneous magnetic field produced by
the current $I$ which flows in the infinite wire at the distance $a$ from the
top rail: $R = 500m\Omega=0.5\Omega ,I = 10A,l = 8.67cm=0.0867m,a =
10.5mm=0.0105m,v =5.36\frac{m}{s}$
\begin{enumerate}
\item What is the electromagnetic force on the rod $\varepsilon _{\tbox{ind}}?$
\item What is the current in the circuit $I_{c}?$
\item What is the rate at which the energy is created $\frac{{dU}}{{dt}}?$
\item What is the external force $F$ needed to keep the velocity $v$ constant?
\item What is the rate of the work of the external force $\frac{dW}{dt}$?
\item Would there still be an induction if the rails don't exist?
\end{enumerate}
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\Dn
{\bf The solution:}
\Dn
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We denote $\kappa=\frac{\mu_0}{4\pi}=\frac{1.26\cdot10^{-6}}{4\pi}
=1.0026\cdot10^{-7}$
\Dn
1. The electro magnetic induction is given by:
%
\begin{eqnarray}
\varepsilon _{\tbox{ind}} &=& -\dot \phi \\
%
\phi &=& \int\limits_a^{a + l} {(vt + x_0)\frac{{2\kappa I}}{r}dr}
= (vt + x_0)\ln \left( {\frac{{a + l}}{a}} \right)2\kappa I
\end{eqnarray}
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where $x_0$ is the initial position of the rod.
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\begin{eqnarray}
\varepsilon _{\tbox{ind}} = -\dot \phi
= -v\ln \left( {\frac{{a + l}}{a}} \right)2\kappa I=2.39\cdot10^{-5}\ Volt
\end{eqnarray}
\Dn
2. The current in the circuit is given by:
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\begin{eqnarray}
I_c = \frac{{\varepsilon _{\tbox{ind}} }}{R} = \frac{{v\ln \left( {\frac{{a +
l}}{a}} \right)2\kappa I}}{R}=4.78\cdot10^{-5}A
\end{eqnarray}
%
in the clockwise direction.
\Dn
3. The rate that the energy is created:
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\begin{eqnarray}
P =\frac{{dU}}{{dt}} = \varepsilon _{\tbox{ind}} \cdot I_c = \frac{{\left[
{v\ln \left( {\frac{{a + l}}{a}} \right)2\kappa I} \right]^2
}}{R}=1.14\cdot10^{-9}watt
\end{eqnarray}
\Dn
4. The external force needed to keep the velocity $v$ constant is:
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\begin{eqnarray}
F=\int\limits_a^{a + l} {dF} = F=\int\limits_a^{a + l} {I_c B(r) dr} =
\int\limits_a^{a + l} {I_c \cdot \frac{{2\kappa I}}{r}dr} = \frac{{\left[ {\ln
\left( {\frac{{a + l}}{a}} \right)2\kappa I} \right]^2 v}}{R}=2.13\cdot10^{-10}N
\end{eqnarray}
\Dn
5. The rate of the work of the external force is:
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\begin{eqnarray}
\frac{{dW}}{{dt}} = \frac{{F \cdot dr}}{{dt}} = F \cdot v = \frac{{\left[ {v\ln \left( {\frac{{a + l}}{a}} \right)2\kappa I} \right]^2 }}{R}=1.14\cdot10^{-9}watt
\end{eqnarray}
and we see that the answer is identical to the answer of q.3.
\Dn
6. Even if there are no rails, consider a charge in the rod during its
movement. The charge ``feels`` the Lorentz force pushing it to the edge of the
rod. Since the charges with the different signs move to the opposite edges of
the rod, it creates voltage between them. It is the electromagnetic induction.
From another point of view we can think of some loop with one of its sides
coincides with the rod and moves and the others are fixed. Then the magnetic
flux through the loop changes with time which creates the elecromagnetic
induction.
But anyway, since the there is no closed circuit here, there is no current.
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