Question

A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C exists along the +x axis. A magnetic field also exists, and its x and y components are Bx = +1.9 T and By = +1.9 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.

Answer 1

a)

`F_(E_x)=1.19\cdot 10^(-3)N` (+x axis)

`F_(B_x)=0`

`F_(B_y)=0`

b)

`F_(E_x)=1.19\cdot 10^(-3) N` (+x axis)

`F_(B_x)=0`

`F_(B_y)=3.21\cdot 10^(-3)N` (+z axis)

c)

`F_(E_x)=1.19\cdot 10^(-3) N` (+x axis)

`F_(B_x)=3.21\cdot 10^(-3) N` (+y axis)

`F_(B_y)=3.21\cdot 10^(-3)N` (-x axis)

Step-by-step explanation:

a)

The electric force exerted on a charged particle is given by

`F=qE`

where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

`q=+4.9\mu C=+4.9\cdot 10^(-6)C` is the charge

`E_x=+242 N/C` is the electric field, along the x-direction

So the electric force (along the x-direction) is:

`F_(E_x)=(4.9\cdot 10^(-6))(242)=1.19\cdot 10^(-3) N`

towards positive x-direction.

The magnetic force instead is given by

`F=qvB sin \theta`

where

q is the charge

v is the velocity of the charge

B is the magnetic field

`\theta` is the angle between the directions of v and B

Here the charge is stationary: this means
`v=0`, therefore the magnetic force due to each component of the magnetic field is zero.

b)

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

`F_(E_x)=1.19\cdot 10^(-3) N` (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

-
`B_x`: this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore,
`\theta=0^(\circ)`, so the force due to this field is zero.

`- B_y`: this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore,
`\theta=90^(\circ)`. Therefore,
`\theta=90^(\circ)`, so the force due to this field is:

`F_(B_y)=qvB_y`

where:

`q=+4.9\cdot 10^(-6)C` is the charge

`v=345 m/s` is the velocity

`B_y = +1.9 T` is the magnetic field

Substituting,

`F_(B_y)=(4.9\cdot 10^(-6))(345)(1.9)=3.21\cdot 10^(-3) N`

And the direction of this force can be found using the right-hand rule:

- Index finger: direction of the velocity (+x axis)

- Middle finger: direction of the magnetic field (+y axis)

- Thumb: direction of the force (+z axis)

c)

As in part b), the electric force has not change, since it does not depend on the veocity of the particle:

`F_(E_x)=1.19\cdot 10^(-3)N` (+x axis)

For the field
`B_x`, the velocity (+z axis) is now perpendicular to the magnetic field (+x axis), so the force is

`F_(B_x)=qvB_x`

And by substituting,

`F_(B_x)=(4.9\cdot 10^(-6))(345)(1.9)=3.21\cdot 10^(-3) N`

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+x axis)

- Thumb: force (+y axis)

For the field
`B_y`, the velocity (+z axis) is also perpendicular to the magnetic field (+y axis), so the force is

`F_(B_y)=qvB_y`

And by substituting,

`F_(B_y)=(4.9\cdot 10^(-6))(345)(1.9)=3.21\cdot 10^(-3) N`

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+y axis)

- Thumb: force (-y axis)