Horizontal wire -- current
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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 an unknown current due to north. Calculate this current knowing that the resulting magnetic field pqcm due to west of the wire is pq.T poing upward at unknown angle above the horizontal and the Earth's field there pos north but downward degree below the horizontal and has magnitude pq.T.
Solution:
The horizontal component of Earth's magnetic field is: B_h B_EarthIndex cosgrad pq.T Since the magnetic field of the horizontal wire is vertical the horizontal component of the resulting magnetic field is also B_hpq.T. With this in mind we can calculate the angle of the resulting magnetic field: alpha arccosleftfrac..right .grad The vertical component of Earth's field and the resulting field are B_ B_EarthIndex singrad pq.T B_ B_textscriptsize res sin.grad pq.T Therefore the magnetic field from the wire is pq.T. The current necessary is therewith I fracpi r Bmu_ pq.A.
A very long horizontal cable carries an unknown current due to north. Calculate this current knowing that the resulting magnetic field pqcm due to west of the wire is pq.T poing upward at unknown angle above the horizontal and the Earth's field there pos north but downward degree below the horizontal and has magnitude pq.T.
Solution:
The horizontal component of Earth's magnetic field is: B_h B_EarthIndex cosgrad pq.T Since the magnetic field of the horizontal wire is vertical the horizontal component of the resulting magnetic field is also B_hpq.T. With this in mind we can calculate the angle of the resulting magnetic field: alpha arccosleftfrac..right .grad The vertical component of Earth's field and the resulting field are B_ B_EarthIndex singrad pq.T B_ B_textscriptsize res sin.grad pq.T Therefore the magnetic field from the wire is pq.T. The current necessary is therewith I fracpi r Bmu_ pq.A.