{{tag>electric_field magnetic_field exam_ee2_SS2024}}{{include_n>1040}} #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Fields of an coax Cable\\ (written test, approx. 12 % of a 120-minute written test, SS2024) #@TaskText_HTML@# A $0.5 ~\rm m$ long coax cable is used for signal transmission. The diagram shows the cross-section of the coax cable with the origin in the center of the coax cable. Due to the given load, the following situation appears: * Inner conductor: $+3.3 ~\rm mA$, $+10 ~\rm nC$ (current into the plane of the diagram) * Outer conductor: $-3.3 ~\rm mA$, $ 0 ~\rm nC$ (current out of the plane of diagram) {{drawio>ee2:ddjurcpk494go2q1_question1.svg}} 1. What is the magnitude of the magnetic field strength $H$ at $\rm (0.1 ~mm | 0)$ and $\rm (0.55 ~mm | 0)$? #@HiddenBegin_HTML~ddjurcpk494go2q1_11,Path~@# The magnitude of the magnetic field strength $H$ can be calculated by: $H = {{I}\over{2 \pi \cdot r}} $ \\ So, we get for $H_{\rm i}$ at $\rm (0.1 ~mm | 0)$, and $H_{\rm o}$ at $\rm (0.55 ~mm | 0)$: \begin{align*} H_{\rm i} &= {{I}\over{2 \pi \cdot r_{\rm i}}} \\ &= {{+3.3 A}\over{2 \pi \cdot { 0.1 \cdot 10^{-3}~\rm m}}} \\ H_{\rm o} &= {{I}\over{2 \pi \cdot r_{\rm o}}} \\ &= {{+3.3 A}\over{2 \pi \cdot { 0.55 \cdot 10^{-3}~\rm m}}} \\ \end{align*} Hint: For the direction, one has to consider the right-hand rule. By this, we see that the $H$-field on the right side points downwards. \\ Therefore, the sign of the $H$-field is negative. \\ But here, only the magnitude was questioned! #@HiddenEnd_HTML~ddjurcpk494go2q1_11,Path~@# #@HiddenBegin_HTML~ddjurcpk494go2q1_12,Result~@# * for $(0.1 ~/rm mm | 0)$ : $H_{\rm i} = 5.25... ~\rm A/m$ * for $(0.55 ~/rm mm| 0)$ : $H_{\rm o} = 0.955... ~\rm A/m$ #@HiddenEnd_HTML~ddjurcpk494go2q1_12,Result~@# 2. Plot the graph of the magnitude of $H(x)$ with $x \in \rm [-0.6~mm, +0.6~mm]$ from $\rm (-0.6 ~mm | 0)$ to $\rm (0.6 ~mm | 0)$ in one diagram. Use proper dimensions and labels for the diagram! #@HiddenBegin_HTML~ddjurcpk494go2q1_21,Path~@# * In general, the $H$-field is proportional to ${{1}\over{r}}$ for the situation between both conductors. * For $H(x)$ with $x$ within the inner conductor, one has to consider how much current is conducted within a circle with the radius $x$. \\ This is proportional to the area within this radius. Therefore, Ther formula $H = {{I}\over{2 \pi \cdot r}} $ gets $H(x) = {{I}\over{2 \pi \cdot x}} \cdot {{\pi x^2}\over{\pi (0.1~\rm mm)^2}}$. This leads to a formula proportional to $x$. * For $x$ within the outer conductor one also gets a linear proportionality with a similar approach. #@HiddenEnd_HTML~ddjurcpk494go2q1_21,Path~@# #@HiddenBegin_HTML~ddjurcpk494go2q1_22,Result~@# {{drawio>ee2:ddjurcpk494go2q1_answer1.svg}} #@HiddenEnd_HTML~ddjurcpk494go2q1_22,Result~@# 3. What is the magnitude of the electric displacement field $D$ at $\rm (-0.1 ~mm | 0)$ and $\rm (0.55 ~mm | 0)$? #@HiddenBegin_HTML~ddjurcpk494go2q1_31,Path~@# The magnitude of the electric displacement field $D$ can be calculated by: $\int D {\rm d}A = Q$. \\ Here, for any position radial to the center, the surrounding area is the surface of a cylindrical shape (here for simplicity without the round endings). \\ This leads to: \begin{align*} D(x) &= {{Q}\over{A}} \\ &= {{Q}\over{l \cdot 2\pi \cdot x}} \\ \end{align*} So, we get for $D_{\rm i}$ at $\rm (0.1 ~mm | 0)$, and $D_{\rm o}$ at $\rm (0.55 ~mm | 0)$: \begin{align*} D_{\rm i} &= {{Q }\over{2 \pi \cdot r_{\rm i} \cdot l}} \\ &= {{10 \cdot 10^{-9} C}\over{2 \pi \cdot { 0.1 \cdot 10^{-3}~\rm m} \cdot 0.5 ~\rm m}} \\ D_{\rm o} &= {{Q }\over{2 \pi \cdot r_{\rm o} \cdot l}} \\ &= {{10 \cdot 10^{-9} C}\over{2 \pi \cdot { 0.55 \cdot 10^{-3}~\rm m}\cdot 0.5 ~\rm m}} \\ \end{align*} Hint: For the direction, one has to consider the sign of the enclosed charge. By this, we see that the $D$-field is positive. \\ But here, again only the magnitude was questioned! #@HiddenEnd_HTML~ddjurcpk494go2q1_31,Path~@# #@HiddenBegin_HTML~ddjurcpk494go2q1_32,Result~@# * for $(0.1 ~\rm mm | 0)$ : $D_{\rm i} = 31.8... ~\rm uC/m^2$ * for $(0.55 ~\rm mm| 0)$ : $D_{\rm o} = 5.78... ~\rm uC/m^2$ #@HiddenEnd_HTML~ddjurcpk494go2q1_32,Result~@# 4. Plot the graph of the magnitude of $D(x)$ with $x \in \rm [-0.6~mm, +0.6~mm]$ from $\rm (-0.6 ~mm | 0)$ to $\rm (0.6 ~mm | 0)$ in one diagram. Use proper dimensions and labels for the diagram! #@HiddenBegin_HTML~ddjurcpk494go2q1_41,Path~@# * In general, the $D$-field is proportional to ${{1}\over{r}}$ for the situation between both conductors. * Since the charges are on the surface of the conductor, there is no $D$-field within the conductor. #@HiddenEnd_HTML~ddjurcpk494go2q1_41,Path~@# #@HiddenBegin_HTML~ddjurcpk494go2q1_42,Result~@# {{drawio>ee2:ddjurcpk494go2q1_answer2.svg}} #@HiddenEnd_HTML~ddjurcpk494go2q1_42,Result~@# #@TaskEnd_HTML@#