Horizontal wire -- magnetic field
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The following quantities appear in the problem:
elektrische Stromstärke \(I\) / Magnetische Flussdichte \(B\) / Winkel \(\theta\) /
The following formulas must be used to solve the exercise:
\(\cos\alpha = \dfrac{b}{c} \quad \) \(B = \dfrac{\mu_0 I}{2\pi r} \quad \) \(a^2+b^2=c^2 \quad \)
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Exercise:
A very long horizontal cable carries .A of current due to north. What is the resulting magnetic field .in due to west of the wire if the Earth's field there pos north but downward ang below the horizontal and has magnitude microtesla?
Solution:
newqtyI.A newqtyro.in solqtyrron*.m newqtypdegree newqtyBeomicrotesla newqtyBeBeon tesla newqtymuo.newtonpersquareampere % prvminipagecm Geg I I r ro r sscphiI p B_EarthIndex Beo Be % GesMagnetic Flux DensityB siT % View from above: center tikzpicturelatex draw nodebelowS -- ++ coordinate N nodeaboveN; drawvery thick darkgreen- noderightI -- ++.; nodeleft at -. W; noderight at . O; drawdarkgreen snakebrace -- nodebelowr ++-. coordinate Bw; fillorange yshiftmmBw nodeleft xshift-mmsscBw circle pt; draworange yshiftmmBw circle pt; fillorange yshiftmmBw nodeleft xshift-mmsscBw circle pt; draworange yshiftmmBw circle pt; fillorange yshiftmmBw nodeleft xshift-mmsscBw circle pt; draworange yshiftmmBw circle pt; nodeorange yshiftmm at Bw largetimes; draworange dashed yshiftmmBw nodeleft xshift-mmsscBperp circle pt; nodeorange yshiftmm at Bw largetimes; draworange dashed yshiftmmBw nodeleft xshift-mmsscBperp circle pt; nodeorange yshiftmm at Bw largetimes; draworange dashed yshiftmmBw nodeleft xshift-mmsscBperp circle pt; draw- orange dashed xshiftmm yshiftmmBw -- ++ noderightsscBh; tikzpicture center View from south: center tikzpicturelatex drawdarkgreen snakebrace -. -- nodebelowr ++-. coordinate Bw; nodedarkgreen at largetimes; drawdarkgreen noderightxshiftmmI circle pt; nodeleft at -. W; noderight at . O; fillorange Bw circle pt; draworang Bw -- ++ nodeabovesscBw; draworang dashed Bw -- ++-. nodebelowsscBperp; nodeorange xshiftmm yshiftmm at Bw largetimes; draworange dashed xshiftmmyshiftmmBw noderight xshiftmmyshiftmmsscBh circle pt; tikzpicture center % The magnetic field due to the wire is solqtyBwfracmu_ Ipi rmuon*In/*pi*rnT al sscBw Bwf fracmuo Ipi r Bw poing upwards. The horizontal component of the resulting field is equal to the horizontal component of the Earth's magnetic field i.e.: solqtyBhB_EarthIndex cossscphiIBen*cosdpnT al sscBh Bhf Be cosp Bh. The vertical component of the resulting field is the Earth's vertical component minus the magnetic field due to the wire i.e.: solqtyBvB_EarthIndex sinsscphiI - fracmu_ Ipi rBen*sindpn-BwnT al sscBV B_perp - sscBw Bvf Be sinp - Bw Bv. Hence the magnitude of the resulting field is according to Pythagoras: solqtyBsqrtB_EarthIndex^ - fracmu_IB_EarthIndexsinsscphiIpi r + fracmu_^I^pi^ r^sqrtBhn**+Bvn**T al B sqrtsscBH^ + sscBV^ prv sqrtB_EarthIndex^cossscphiI + B_EarthIndex^sinsscphiI - fracmu_IB_EarthIndexsinsscphiIpi r + fracmu_^I^pi^ r^ prv Bf sqrtqtyBh^ + qtyBv^ B. B Bf BII TecB- prvminipage
A very long horizontal cable carries .A of current due to north. What is the resulting magnetic field .in due to west of the wire if the Earth's field there pos north but downward ang below the horizontal and has magnitude microtesla?
Solution:
newqtyI.A newqtyro.in solqtyrron*.m newqtypdegree newqtyBeomicrotesla newqtyBeBeon tesla newqtymuo.newtonpersquareampere % prvminipagecm Geg I I r ro r sscphiI p B_EarthIndex Beo Be % GesMagnetic Flux DensityB siT % View from above: center tikzpicturelatex draw nodebelowS -- ++ coordinate N nodeaboveN; drawvery thick darkgreen- noderightI -- ++.; nodeleft at -. W; noderight at . O; drawdarkgreen snakebrace -- nodebelowr ++-. coordinate Bw; fillorange yshiftmmBw nodeleft xshift-mmsscBw circle pt; draworange yshiftmmBw circle pt; fillorange yshiftmmBw nodeleft xshift-mmsscBw circle pt; draworange yshiftmmBw circle pt; fillorange yshiftmmBw nodeleft xshift-mmsscBw circle pt; draworange yshiftmmBw circle pt; nodeorange yshiftmm at Bw largetimes; draworange dashed yshiftmmBw nodeleft xshift-mmsscBperp circle pt; nodeorange yshiftmm at Bw largetimes; draworange dashed yshiftmmBw nodeleft xshift-mmsscBperp circle pt; nodeorange yshiftmm at Bw largetimes; draworange dashed yshiftmmBw nodeleft xshift-mmsscBperp circle pt; draw- orange dashed xshiftmm yshiftmmBw -- ++ noderightsscBh; tikzpicture center View from south: center tikzpicturelatex drawdarkgreen snakebrace -. -- nodebelowr ++-. coordinate Bw; nodedarkgreen at largetimes; drawdarkgreen noderightxshiftmmI circle pt; nodeleft at -. W; noderight at . O; fillorange Bw circle pt; draworang Bw -- ++ nodeabovesscBw; draworang dashed Bw -- ++-. nodebelowsscBperp; nodeorange xshiftmm yshiftmm at Bw largetimes; draworange dashed xshiftmmyshiftmmBw noderight xshiftmmyshiftmmsscBh circle pt; tikzpicture center % The magnetic field due to the wire is solqtyBwfracmu_ Ipi rmuon*In/*pi*rnT al sscBw Bwf fracmuo Ipi r Bw poing upwards. The horizontal component of the resulting field is equal to the horizontal component of the Earth's magnetic field i.e.: solqtyBhB_EarthIndex cossscphiIBen*cosdpnT al sscBh Bhf Be cosp Bh. The vertical component of the resulting field is the Earth's vertical component minus the magnetic field due to the wire i.e.: solqtyBvB_EarthIndex sinsscphiI - fracmu_ Ipi rBen*sindpn-BwnT al sscBV B_perp - sscBw Bvf Be sinp - Bw Bv. Hence the magnitude of the resulting field is according to Pythagoras: solqtyBsqrtB_EarthIndex^ - fracmu_IB_EarthIndexsinsscphiIpi r + fracmu_^I^pi^ r^sqrtBhn**+Bvn**T al B sqrtsscBH^ + sscBV^ prv sqrtB_EarthIndex^cossscphiI + B_EarthIndex^sinsscphiI - fracmu_IB_EarthIndexsinsscphiIpi r + fracmu_^I^pi^ r^ prv Bf sqrtqtyBh^ + qtyBv^ B. B Bf BII TecB- prvminipage