Question

A very long, straight wire carries a current of 18.3 A out of the screen. An electron outside the wire is 1.79 cm to the right of the central axis of the wire and is moving with a speed of 5.95×10⁶ m/s.

A. Find the magnitude of the magnetic force on the electron if it is moving directly away from the wire (i.e., rightward).

B. Find the magnitude of the force on the electron if it is moving parallel to the wire in the direction of the current (i.e., out of the screen)

C.Find the magnitude of the force on the electron if it is moving upward

Answer 1

A)

r = distance of electron from wire = 1.79 cm = 0.0179 m

i = current in the long wire = 18.3 A

B = magnetic field at the location of the electron

Magnetic field at the location of the electron is given as

B = (/4) (2i/r)

B = (10^-7) (2(18.3)/0.0179)

B = 204.5 x 10^-6 T                           in upward direction

v = velocity of electron = 5.95 x 10⁶ m/s                        in right direction

q = charge on electron = - 1.6 x 10^-19 C

Magnitude of magnetic force on the electron is given as

F = q (v x B )

F = q v B Sin90

F = (1.6 x 10^-19) (5.95 x 10⁶) (204.5 x 10^-6) Sin90

F = 1.95 x 10^-16 N

B)

Magnitude of magnetic force on the electron is given as

F = q (v x B )

F = q v B Sin90

F = (1.6 x 10^-19) (5.95 x 10⁶) (204.5 x 10^-6) Sin90

F = 1.95 x 10^-16 N

c)

Magnitude of magnetic force on the electron is given as

F = q (v x B )

F = q v B Sin0

F = (1.6 x 10^-19) (5.95 x 10⁶) (204.5 x 10^-6) Sin0

F = 0 N