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
?
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?
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.
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,
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:
).
The expression for electric force
is given by,

Here
is the first point charge,
is the second point charge,
is the distance between the two charges and
is the permittivity of free space (Value:
) .
Magnetic force equation is given by,
Here,
is the charge ,
is the velocity of the charge,
is the magnetic field and
is the angle between
and
.
The expression of the net magnetic force
on each charge exerts on another can be expressed as,

Here,
and
are the charges,
and
are the velocities of the charges and
is the distance between the two charges.
The magnitude of magnetic field
for the first charge is given by,
Substituting
for
,
for
,
for
,
for
and
.
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The magnetic field
for the second charge is given by,
Substituting
for
,
for
,
for
,
for
and
for
.
The magnitude of the net magnetic field
due to two charges at point P can be calculated as,

Substitute
for
and
for
.

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

Substituting
for
,
for
,
for
,
for
and
for
.
[Part B]
Part B
Part B
The net magnetic force that each charge exerts each other can be calculated as,

Substitute
for
,
for
,
for
,
for
,
for
and
for
.

[Part C]
Part C
Part C
Calculate the ratio R by dividing the magnitude of electric force and the magnetic force.
Substitute
for
and
for
.
[Part D]
Part D
Ans: Part AThe magnitude of the net magnetic field at point P is
T.
28.6)Positive point charges q = 9.00 μC andq′= 4.00 μCare moving relative to an observer at point P, as shown in the figure (Figure 1) . The distanced is 0.130 m , v = 4.40×106 m/s , andv′= 8.80×106 m/s .If the direction of v⃗ ′ is reversed, so both charges are moving in the same direction, what is the magnitude of the magnetic forces that the two charges exert on each other?
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Exercise 28.6
Positive point charges q = 7.00 ?C and
q?= 4.00 ?C are moving relative to an observer at
point P, as shown in the figure (Figure 1) . The distance
d is 0.130 m , v = 4.40×106 m/s , and
v?= 8.80×106 m/s .
Part H
If the direction of v? ? is reversed, so both charges
are moving in the same direction, what is the magnitude of the
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