A particle has a charge of q 53 mu C and is located at the

A particle has a charge of q = +5.3 mu C and is located at the origin. As the drawing shows, an electric field of E_x = +211 N/C exists along the +x axis. A magnetic field also exists, and its x and y components are B_x = + 1.6 T and B_y = + 1.6 T. Calculate the force (magnitude and direction) exerted on the particle by each of the three fields when it is stationary, moving along the +x axis at a speed of 345 m/s, and moving along the +z axis at a speed of 345 m/s.

Solution

Q= +5.3 C

E_x = +211 N/C

B_x = 1.6 T

B_y = 1.6 T

when the particle is stationary it will not experience any force by the magnetic field, the only force on it is due to the Electric field.

F_E = E_x*Q = 211*5.3e-6 = 1.12 e-3 N in +x direction

When the particle has a speed of 345 m/s +x –direction

it is perpendicular to the magnetic field B_y and parallel to B_x

Force on a moving charge due to magnetic field is

qv xB , hence force due to B_x is 0

Force due to B_y

F_M = 5.3e-6*345*1.6 = 2.93e-3 N , the direction is +z axis

F_E   = 1.12 e-3 N in +x direction

F = 1.12e-3 x^ + 2.93 z^

magnitude of F = 3.13e-3 N

it makes +63.97⁰ with +x in xz plane

When the particle is moving in +z at a speed of 345m/s

the speed is perpendicular to each of the fields B_x and B_y and the magnitude of the force due to each is

3.13e-3 N and is perpendicular to the field and the speed.

direction of force due to B_x = v x B = k X i = j

direction of force due to B_y = k X j = -i

F = 1.12e-3 x^ -3.13 x^ +3.13 y^

    = -2.01e-3 x^ +3.13 y^

Magnitude of F = 3.72 N

it makes +57.29⁰ with –x-axis in x-y plane