Question

A long straight wire carries a current of 40 A to the right. An electron, traveling at 2.7 10 7 m/s, is 5.3 cm from the wire. What force, magnitude and direction, acts on the electron if the electron velocity is directed toward the wire

Answer 1

Answer: f = 6.52*10^-16 N

Step-by-step explanation:

if we assume that the force is directed at the y positive direction, then

B = μi / 2πr, where

μ = 4π*10^-7

B= (4π*10^-7 * 40) / 2 * π * 5.3*10^-2

B = 5.027*10^-5 / 0.333

B = 1.51*10^-4 T

Since v and B are perpendicular, then,

F = qvB

F = 1.6*10^-19 * 2.7*10^7 * 1.51*10^-4

F = 2.416*10^-23 * 2.7*10^7

F = 6.52*10^-16 N

Therefore, the magnitude of the force is, F = 6.52*10^-16 N and it moves in the i negative direction

Answer 2

`6.53*10^-^1^7N`

Step-by-step explanation:

The magnet of the magnetic field is 53 cm = 0.53m from wire is

`B = (\mu_0 I)/(2\pi d)`

`= ((4\pi * 10^-^7)(40))/(2 \pi (0.53)) \\\\= (5.0265* 10^-^5)/(3.33) \\\\= 1.5095 * 10^-^5`

the magnetic force exerted by the wire on the electron is

`F = Bqv \sin \theta\\\\= 1.5095 * 10^-^5 *1.602*10^-^1^9*2.7*10^7\\\\= 6.53*10^-^1^7N`

From the right hand rule the direction of the force is parallel to the current (since the particle is electron)