Tube!

a) Use the equation for the resistance in terms of its resistivity and its dimensions.

b) Use Ohm’s Law to solve for the current.

c) Find the velocity using the current in terms of the density and the charge. Find the time for an object moving at constant speed.

a) The resistance of a conducting wire is given by

\begin{equation*}
R = \frac{\rho L}{A},
\end{equation*}

where the area for a cross-sectional circle is \(A = \pi r^2\). Then, with numerical values:

\begin{equation*}
R = 5.47\times10^9 \ \Omega.
\end{equation*}

b) By Ohm’s law, solving for \(I\) we get:

\begin{equation*}
I = 7.67\times10^{-8} \ \text{A}.
\end{equation*}

c) The current flow produced by the flow of charged particles is:

\begin{equation*}
I = qnAv,
\end{equation*}

where \(n\) is the free charge density. The time it takes for a particle to travel a given distance with constant speed is:

\begin{equation*}
t = \frac{L}{v}.
\end{equation*}

Solving for the velocity in the current equation and replacing it in the time we get:

\begin{equation*}
t = \frac{LenA}{I},
\end{equation*}

or with numerical values:

\begin{equation*}
t = 1.41\times10^{17} \ \text{s}.
\end{equation*}

For a more detailed explanation of any of these steps, click on “Detailed Solution”.

a) The figure shows the dimensions of the tube. We should find the resistance by relating it to the resistivity and the dimensions of the tube.

The tube is made of metal, which is a conductor. The resistance of a conducting wire is given by

\begin{equation}
\label{EQ:1}
R = \frac{\rho L}{A},
\end{equation}

where \(\rho\) is the resistivity, \(L\) is the length, and \(A\) is the cross-sectional area of the wire. For a cylindrical tube, the cross-sectional area is given by

\begin{equation}
A = \pi r^2.
\end{equation}

After substituting this in eq. \eqref{EQ:1}, the resistance becomes:

\begin{equation}
R = \frac{\rho L}{\pi r^2},
\end{equation}

and after inserting numerical values, we get

\begin{equation}
R = \frac{1.72\times10^8 \ \ \Omega \text{ m}\cdot(4 \ \text{m})}{3.1416\cdot(0.2 \text{ m})^2}
= 5.47\times10^9 \ \Omega.
\end{equation}

b) We are given the voltage at which the tube is subjected to, and we have just found its resistance; hence, we can use Ohm’s law to find the current across the tube.

Ohm’s law states that the voltage \(V\) at which a material is subjected, the current across it \(I\), and its resistance \(R\) are related by the equation

\begin{equation}
V = I R.
\end{equation}

If we divide by R into both sides, we get

\begin{equation}
I = \frac{V}{R},
\end{equation}

and if we insert numerical values, we obtain

\begin{equation}
I = \frac{420 \text{ V}}{5.47\times10^9 \ \Omega}
= 7.67\times10^{-8} \ \text{A}.
\end{equation}

c) Current is produced by the motion of charges, and it is defined as the charge per unit time that travels across the wire. In this case (as in all metals), these charges are electrons that are free to circulate across the metal when a voltage is applied to it. The density of these free electrons was given to us. Therefore, in order to find the time an electron takes to cross the tube, first, we need to relate the current with the dimensions of the tube, the charge of the electron, and the density of free electrons to find the average speed of the electrons. Then, we can find the time by using this speed and the length of the tube.

The current \(I\) produced by the flow of charged particles with charge \(q\), moving with speed v, on a conductor of having a cross-sectional area \(A\), and a free charge density \(n\) is given by.

\begin{equation}
I = qnAv.
\end{equation}

If we divide by \(qnA\) on both sides of the equation, we get

\begin{equation}
v = \frac{I}{qnA}.
\end{equation}

In this case, the moving charges are electrons, and thus \(q = -e\). Therefore,

\begin{equation}
v = -\frac{I}{enA}.
\end{equation}

The sign indicates in which direction the electrons are moving relative to the direction of the current. Here, we only care about the magnitude of the velocity. Hence, if we take the absolute value, we get

\begin{equation}
\label{EQ:2}
|v| = \frac{I}{enA}.
\end{equation}

Now, the speed \(|v|\) is defined as the distance traveled per unit time. In this case, we can write it as the length of the tube \(L\) divided by the time \(t\) it takes for an electron to cross the tube:

\begin{equation}
|v| = \frac{L}{t},
\end{equation}

which we can rewrite after multiplying on both sides by \(\frac{t}{v}\) as

\begin{equation}
t = \frac{L}{|v|}.
\end{equation}

After substituting \(|v|\) from eq. \eqref{EQ:2}, we obtain

\begin{equation}
t = \frac{L}{\frac{I}{enA}}
= \frac{LenA}{I},
\end{equation}

and if we insert numerical values, we get

\begin{equation}
t = \frac{(4 \text{ m})\cdot(1.6\times10^{-19} \ \text{C})\cdot(8.5\times10^{28} \ \frac{1}{\text{m}^3} )*(0.2 \ \text{m})}{7.67\times10^{-8} \ \text{A}}
= 1.41\times10^{17} \ \text{s}.
\end{equation}

The units of time in this case are seconds because \(1 \ \text{A} = \ 1 \ \frac{\text{C}}{\text{s}}.\)

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