A long straight conducting wire carries a current I = 3.5 A to the left. An elec

Question

A long straight conducting wire carries a current I = 3.5 A to the left. An electron (with negative charge -e) is moving with a constant velocity upsilon = 1.7 times 10^6 m/s to the left below the wire. The distance between the electron and the wire is d = 4.6 cm. There is also an electric field in this region, but we don't know its magnitude and direction. (Ignore the gravity of the electron.) (a) Find the magnitude AND direction of the magnetic field produced by the current at the electron's location. (b) Find the magnitude AND direction of the magnetic force on the electron. (c) The electron moves with a constant velocity under both the magnetic and electric forces. Find the magnitude AND direction of the electric field in this region.

Explanation / Answer

a) magnetic field produce by wire = B = Uo*i/(2*pi*d) = 4*pi*10^-7*3.5/(2*pi*0.046) = 152.17*10^-7 T

direction - out of the page

b) magnitude of magnetic force on the electron F= e*v*B = 1.602*10^-19*1.7*10^6*152.17*10^-7 = 414.41*10^-20 N

direction is away from the wire

c) electron moves with constant velocity

=> electric force = magnetic force

=> e*E = F = 414.41*10^-20

=> E = 414.41*10^-20/(1.602*10^-19) = 25.86 N/C away from the wire .

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