Question

A uniform magnetic field of magnitude 0.80 T in the negative z-direction is present in a region of space. A uniform electric field is also present. In Figure 20.3, theelectric field is set at 42,400 V/m in the positive y-direction. An electron is projected with an initial velocity vo = 5.3 × 104 m/s in the positive x-direction. They-component of the initial force on the electron is closest to:

Answer 1

The force from the magnetic field is found from the formula F = qvB

F = (1.6 X 10^-19)(5.3 X 10⁴)(.8) = 6.784 x 10^-15 N in the negative y direction (from the right hand rule)

The force from the electric field is found from the formula F = qE

F = (1.6 X 10^-19)(42400) = 6.784 X 10-15 N in the negative y direction (E field point up for a positive charge, thus is pulls down for a negative charge.

Since we have two negative y forces, the sum is 6.784 X 10^-15 + 6.784 X 10^-15 = 1.36 X 10^-14 N

Since it is in the negative direction, this value is closest to your choice C, -1 X 10^-14 N

Answer 2

By the right hand rule:
thumb points in direction of velocity (+x direction)
fingers point in direction of magnetic field (-z direction)
palm point in direction of magnetic force (+y direction)

Since this is an electron, the magnetic force will be opposite to what the right hand rule states:
magnetic force (-y direction)

Since the electric field points in the +y direction, the electron will experience an electric force in the opposite direction:
electric force (-y direction)

Therefore the y_component of the force on the electron will be:
F = FE + FM
F = (Eq) + (qvB)
F = (42400*1.6e-19) + (1.6e-19)(5.3e4)(0.8)
F = 6.78e-15 N

This will be in the -y direction

BOL

Answer 3

by the rule of cross product it is clear that the magnetic force is in the y-direction [ i x (-k) = j ]
and the electric force is also in the +ve y-direction
So initial force = -e[(vB) + E]j = 135680*10^-19N

Answer 4

Lu c Bela to C.

.

Answer 5

take help i solved for diff values put yours
There will be two forces acting on the electron in the y direction. They will be due to both the

1. Magnetic field by the equation F=qv X B=qvBsin(theta)

2. Electric field by the equation F=qE

where
q=1.602e-19 and the rest of the variables are given

1. F=qvBsin(theta)=
1.602e-19*2.6e4*0.8*sin(pi/2)
=3.33e-15 N

2. F= qE= 1. 602e- 19*20800= 3.33e-15 N => -3.33e-15 N

Because the electric field is in the positive y direction and it is influencing a negatively charged electron,


then the force produced by the electric field on the electron will be in the -y direction.

Thus by superposition,

F1+F2=3.33e-15-3.33e-15=0 N