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Tags conductor, electromagnetism, magnetism, physics, terrestrial magnetic field, trigonometric function, vector, vectorgeometry Difficulty Points 2 Language Type Quantity Last modified 2 days, 15 hours ago Initial Version by Urs Zellweger (urs)	Magnetfeld eines geraden Leiters (patrik) Magnetfeld - Gerader Leiter (urs) Quelle des Magnetismus 1 (urs) AG - Horizontal wire (urs)	0

Exercise

A very long horizontal cable carries $\SI{23.4}{A}$ of current due to north. What is the resulting magnetic field $\SI{7.09}{in}$ due to west of the wire, if the Earth's field there points north but downward, $\ang{37}$ below the horizontal, and has magnitude $\SI{48}{\micro\tesla}$? $\star$

Solution

\newqty{I}{23.4}{A} \newqty{ro}{7.09}{in} \solqty{r}{}{\ron*2.54e-2}{m} \newqty{p}{37}{\degree} \newqty{Beo}{48}{\micro\tesla} \newqty{Be}{\Beon e-6}{\tesla} \newqty{muo}{1.256e-6}{\newton\per\square\ampere} % \Geg{ I &= \I \\ r &= \ro = \r \\ \ssc{\phi}{I} &= \p \\ B_{\EarthIndex} &= \Beo = \Be } % \Ges{Magnetic Flux Density}{[B] = \si{T}} % The magnetic field due to the wire is \solqty{Bw}{\frac{\mu_0 I}{2\pi r}}{\muon*\In/(2*pi*\rn)}{T} \al{ \ssc{B}{w} &= \Bwf \\ &= \frac{\muo \cdot \I}{2\pi \cdot \r} \\ &= \Bw } pointing upwards. The horizontal component of the resulting field is equal to the horizontal component of the Earth's magnetic field, i.e.: \solqty{Bh}{B_{\EarthIndex} \cdot \cos(\ssc{\phi}{I})}{\Ben*cosd(\pn)}{T} \al{ \ssc{B}{h} &= \Bhf \\ &= \Be \cdot \cos(\p) \\ &= \Bh. } The vertical component of the resulting field is the Earth's vertical component minus the magnetic field due to the wire, i.e.: \solqty{Bv}{B_{\EarthIndex} \cdot \sin(\ssc{\phi}{I}) - \frac{\mu_0 I}{2\pi r}}{\Ben*sind(\pn)-\Bwn}{T} \al{ \ssc{B}{V} &= B_\perp - \ssc{B}{w} = \Bvf \\ &= \Be \cdot \sin(\p) - \Bw \\ &= \Bv. } Hence, the magnitude of the resulting field is according to Pythagoras: \solqty{B}{\sqrt{B_{\EarthIndex}^2 - \frac{\mu_0IB_{\EarthIndex}\sin(\ssc{\phi}{I})}{\pi r} + \frac{\mu_0^2I^2}{4\pi^2 r^2}}}{sqrt(\Bhn**2+\Bvn**2)}{T} \al{ B &= \sqrt{\ssc{B}{H}^2 + \ssc{B}{V}^2} = \sqrt{B_{\EarthIndex}^2\cos[2](\ssc{\phi}{I}) + B_{\EarthIndex}^2\sin[2](\ssc{\phi}{I}) - \frac{\mu_0IB_{\EarthIndex}\sin(\ssc{\phi}{I})}{\pi r} + \frac{\mu_0^2I^2}{4\pi^2 r^2}} = \Bf \\ &= \sqrt{\qty(\Bh)^2 + \qty(\Bv)^2} \\ &= \B. } \Lsg{ B &= \Bf \\ &= \BII = \Tec{B}{2}{-6} }

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