Question

A particle has a charge of q = +5.6 μC and is located at the origin. As the drawing shows, an electric field of Ex = +213 N/C exists along the +x axis. A magnetic field also exists, and its x and y components are Bx = +1.0 T and By = +1.6 T. Calculate the force (magnitude and direction) exerted on the particle by each of the three fields when it is (a) stationary, (b) moving along the +x axis at a speed of 345 m/s, and (c) moving along the +z axis at a speed of 345 m/s.

(a) FE =

FBx =

FBy =

(b) FE =

FBx =

FBy =

(c) FE =

FBx =

FBy =

Answer 1

Given,

q = 5.6 uC ; Ex = 213 N/C ; Bx = 1 T ; By = 1.6 T

a)When the charge is stationary only electric force acts on it.

FE = q E

FE = 5.6 x 10^-6 x 213 = 1.193 x 10^-3 N

Hence, FE = 1.193 x 10^-3 N ; FBx = 0 ; FBy = 0

b)FE = qE = 1.193 x 10^-3 N

FBx = q v B sin0 = 0 N

FBy = q v B sin90 = q v B

FBy = 5.6 x 10^-6 x 345 x 1.6 = 3.091 x 10^-3 N

Hence, FE = 1.193 x 10^-3 N ; FBx = 0 N ; FBy = 3.091 x 10^-3 N

c)FE = qE = 1.193 x 10^-3 N

FBx = q v B sin90 = q v B

Fbx = 5.6 x 10^-6 x 345 x 1 = 1.932 x 10^-3 N

FBy = q v B sin90 = q v B

FBy = 5.6 x 10^-6 x 345 x 1.6 = 3.091 x 10^-3 N

Hence, FE = 1.193 x 10^-3 N ; FBx = 1.932 x 10^-3 N ; FBy = 3.091 x 10^-3 N