a) [6] A $50 \mathrm{~Hz}$ alternating current of $10 \mathrm{~A}$ (rms) passes along a copper wire whose diameter is $1.0 \mathrm{~mm}$. If copper contains $9.0 \times 10^{22}$ free electrons per $\mathrm{cm}^{3}$, estimate
(i) the maximum speed of the electrons, and
(ii) the maximum displacement of the electrons associated with the current.
b) [6] A wire of length $l$ and resistance $R$ is extended by an amount $\varepsilon$ such that $\varepsilon \ll l$. If the volume of the wire and its resistivity remain unchanged,
(i) Give an expression for the fractional change in $R$, i.e. $\frac{\delta R}{R}$.
(ii) Calculate the new resistance of the wire if $R=100.0 \Omega$ and $\varepsilon / l=0.001$.
c) [6] A non-linear electrical component is connected in a circuit in series with a supply of emf $\varepsilon=300 \mathrm{~V}$ and negligible internal resistance, and a resistor $R$ of $4.0 \mathrm{k} \Omega$. The relation between the current $I$ through the non-linear component, and the potential difference across it, $V_{c}$, is given by $I=A V_{\mathrm{c}}+B V_{\mathrm{c}}^{2}$ where $A=0.070 \mathrm{~mA} \mathrm{~V}{ }^{-1}, B=0.0050 \mathrm{~mA} \mathrm{} \mathrm{V}^{-2}$.
(i) Obtain an algebraic expression for $V_{c}$.
(ii) Determine the current $I$ through the non-linear component.
d) [7] A metal hemisphere of radius $0.10 \mathrm{~m}$ is immersed near the centre of a large conducting tank containing a liquid of resistivity $60 \Omega \mathrm{m}$. The plane surface of the hemisphere is level with the surface of the liquid.
(i) Find an expression for the resistance between two hemispherical shells within the liquid, concentric with the centre of the metal hemisphere, having radii $r$ and $(r+\delta r)$ respectively.
(ii) Calculate the resistance between the hemisphere and the tank.
Show answer · 显示答案
- (a)(i)$\displaystyle v_{\text{max}} = 1.25 \times 10^{-3} \, \text{m/s}$
- (a)(ii)$\displaystyle 3.98 \times 10^{-6} \, \text{m}$
- (b)(i)$\displaystyle 2 \frac{\varepsilon}{l}$
- (b)(ii)$\displaystyle 100.2 \, \Omega$
- (c)(i)$\displaystyle 94.6 \, \text{V}$
- (c)(ii)$\displaystyle 0.0513 \, \text{A}$
- (d)(i)$\displaystyle dR = \frac{\rho \delta r}{2 \pi r^2}$
- (d)(ii)$\displaystyle 95.5 \, \Omega$
Worked solutions are in the book.































