A long, right wire dead a present of 2.5 μA. One electron moves parallel to the wire v a consistent speed that 3.6x104 m/s at street "d" below the wire. What is the magnitude and also direction fo the magnetic ar at the location of the electron?
Hi Gnarls (I favored "Crazy," by the way):
The trick in this type of problem typically is what they carry out not tell you, because it seems favor there might not be enough information. The key, though, is in the expression "parallel come the wire." A charged fragment in a ar should normally curve. If that does not, there should be some other force on the balancing the end the magnetic force. In this case, it is the pressure of gravity on the electron.
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Therefore, the magnetic pressure on the electron needs to be "up" (i.e. Toward the wire, due to the fact that the electron is stated to it is in below). And its magnitude should equal mg for the electron (whose mass deserve to be easily looked up).
Since the electron is traveling parallel come the wire, and the magnetic ar of the cable exists in concentric circles approximately the wire, the particle must be travel perpendicular come the field as it relocate parallel come the cable (by geometry). Therefore, the magnetic pressure on that is qvB. Because v is given and also the charge q of the electron is likewise a typically known worth (easy to look up), this allows solving for the strength of the magnetic ar at the electron"s position.
There is a typical formula because that the magnetic ar a perpendicular street r indigenous a long, straight wire:
B = μoi/2πr, where r is the perpendicular distance and also i is the existing (and μo is (magnetic) permeability (of totally free space), an additional universal constant)
This will permit you to deal with for r, the distance from the wire.
The direction that the present is a selection between "in the same direction as the particle" and "opposite the direction of the particle." You have to use the right-hand ascendancy for the force on a moving charge to identify the direction of the field to gain an upward force on the electron listed below the cable (remember that the electron has actually a an unfavorable electric charge). One you recognize the direction the the field listed below the wire, then identify (by a various right-hand rule) the direction the current needs to go in the cable to acquire the magnetic ar in the ideal direction listed below it.
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For comparison, I gained the street to be 323 m and the direction that the existing to it is in opposite the direction that the electron"s take trip (I perform not make any kind of guarantees about the correctness of this results, yet I confirm my work-related a pair times and also got the very same answer; if ns think of one error later, I will certainly repost).