Horizontal wire -- current
Question
Solution
Short
Video
\(\LaTeX\)
Need help? Yes, please!
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:
\(\cos\alpha = \dfrac{b}{c} \quad \) \(B = \dfrac{\mu_0 I}{2\pi r} \quad \) \(a^2+b^2=c^2 \quad \)
No explanation / solution video to this exercise has yet been created.
Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
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 east of the wire is pq.T poing upward above the horizontal and the Earth's field there pos north degree above the horizontal and has magnitude pq.T.
Solution:
The horizontal component of the magnetic field of the Earth is: B_h B_EarthIndex cosgrad pq.T Since the magnetic field occuring due to the current in the wire is perpicular to Earth's surface i.e. vertical the horizontal component of the resulting field is also B_h which allows us to calculate the angle of the resulting field: alpha arccosleftfrac..right .grad The vertical component of the Earth's field and the resulting magnetic field are B_ B_EarthIndex singrad pq.T B_ B_r singrad pq.T respectively. The field produced by the wire is therefore B B_-B_ pq.T. The current in the wire herewith is: 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 east of the wire is pq.T poing upward above the horizontal and the Earth's field there pos north degree above the horizontal and has magnitude pq.T.
Solution:
The horizontal component of the magnetic field of the Earth is: B_h B_EarthIndex cosgrad pq.T Since the magnetic field occuring due to the current in the wire is perpicular to Earth's surface i.e. vertical the horizontal component of the resulting field is also B_h which allows us to calculate the angle of the resulting field: alpha arccosleftfrac..right .grad The vertical component of the Earth's field and the resulting magnetic field are B_ B_EarthIndex singrad pq.T B_ B_r singrad pq.T respectively. The field produced by the wire is therefore B B_-B_ pq.T. The current in the wire herewith is: I fracpi r Bmu_ pq.A