Question

Magnetic Fields and Forces on Point Charges

Positive point charges q = 9.00 μC and q'= 4.00 μC are moving relative to an observer at point P, as shown in the figure . The distance d is 0.120 m, v = 4.40×106m/s, and v'= 8.80×106 m/s.

Part A:

When the two charges are at the locations shown in the figure, what is the magnitude of the net magnetic field they produce at point P?

Part B:

What is the magnitude of the electric force that each charge exerts on the other?

Part C:

What is the magnitude of the magnetic force that each charge exerts on the other?

Part D:

What is the ratio of the magnitude of the electric force to the magnitude of the magnetic force?

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Answer #1
Concepts and reason

The concepts used to solve this problem are right hand thumb rule, the expression of the magnetic field, electric field, magnetic force and electric force.

First calculate the direction and magnitude of the magnetic field due to moving charge and at given some given point in the given figure.

Calculate the net magnetic field by adding the two magnetic fields keeping their direction in mind. Find the direction using the right-hand thumb rule.

Calculate the magnetic force and the electric force by using their expression and finally calculate their ratio.

Fundamentals

Right Hand Thumb Rule: This rule states that if the thumb of right hand points in the direction of the moving charge, then the fingers will determine the direction of the magnetic field.

The expression for magnetic field is given by,

||B|= 4 gvsin 0
4л г?

Here is the charge for which the magnetic field is to be found, v is the velocity of the moving charge,is the angle made by the line joining the point and the moving charge direction. r is the distance between the two moving charges, while is the permeability of free space (Value:4л х107 т-m-А-
).

The expression for electric force F
elec
is given by,

2
Απεο
elec

Here is the first point charge,42
is the second point charge, is the distance between the two charges and is the permittivity of free space (Value:|8.85x1012 F m
) .

Magnetic force equation is given by,

F
m vBsin

Here, is the charge ,V
is the velocity of the charge,В
is the magnetic field and is the angle between V
and В
.

The expression of the net magnetic force on each charge exerts on another can be expressed as,

F - М. д9vy
4л (г)
2

Here,and are the charges,V
and are the velocities of the charges and is the distance between the two charges.

The magnitude of magnetic field В.
for the first charge is given by,

|В.Е Но дysin 0
4л р?

Substituting 9 HC
for ,4.4x106 m.s
for V
,0.120 m
for ,90°
for and 4л х107 т-m-А-
.

(4л х10Т-m-А-) (9иC)1x10*C\\44x10° ms
|в]-
|В.
(4я)
s1)(sin 90)
1 ДС
(0.120 m)
3D 2.75х104 T

The magnetic field В.
for the second charge is given by,

| B, |= 4, gvsin o

Substituting 4 С
for ,8.8x10 m-s
for V
,0.120 m
for ,90°
for and 4л х107 т-m-А-
for .

1x 10 C8 8x10° m-s
(4rx10 T-m-A)(4C)
(4T)
)(sin90)
1μC
|В.
(0.120m
=2.44x 104 T

The magnitude of the net magnetic field В.
net
due to two charges at point P can be calculated as,

net

Substitute |2.75x10 T
4
for and 2.44x 10 T
for .

B(2.75x10 T)+(2.44x10 T)
net
=5.19x 10 T

The direction of net magnetic field at P and vertically inside the paper.

[Part A]

Part A

Part A

Consider them as point charges, calculate the electric force by substituting values in the formula of the electric force provided in the fundamentals section.

The distance between the charges can be calculated as,

r dd
= 2d

The electric force due to two charges and is given by,

1
Απεο (2d)
elec

Substituting 9 HC
for ,4 АC
for ,0.12 m
for ,90°
for and |8.85x1012 F m
for .

(4uC) lx10c
C
(9uC lx10 c
1
F
(4)(3.14) (8.85x1012 F.m1)
elec
2 (0.12 m)
=5.625 N

[Part B]

Part B

Part B

The net magnetic force that each charge exerts each other can be calculated as,

4л (2d)

Substitute for ,0.120 m
for , for , for ,4.40x 10 m s
for V
and 8.80x10 m-s
for .

(47x107 T m-A
(47
1x10 C
5)(8.80x10 m-s)
(9uC)1x10
4.40x10 m s
1C
2 (0.120m)
=2.42x103 T

[Part C]

Part C

Part C

Calculate the ratio R by dividing the magnitude of electric force and the magnetic force.

|F.
elec
R =
mag

Substitute 5.625 N
for |F
elec
and 2.42x103 T
for mag
.

|Felee 5.625 N
2.42x103T
mag
= 2324.4

[Part D]

Part D

Ans: Part A

The magnitude of the net magnetic field at point P is 4
5.19x10
T.

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