Calculating the Forces on a Charged Particle in an Electric and Magnetic Field

In summary, the magnet produces a 0.40 T field between its poles, directed horizontally to the east. A dust particle with charge q = -8.0×10^-18 C is moving vertically downwards with a speed of 0.30 cm/s. It experiences a magnetic force of 9.6 x 10^-21 N directed north and an electric force of -8 x 10^-20 N directed south. The net force on the dust particle is 7.04 x 10^-20 N.
  • #1
quik
3
0

Homework Statement



A magnet produces a 0.40 T field between its poles, directed horizontally to the east. A dust particle with charge q = -8.0×10^-18 C is moving vertically downwards with a speed of 0.30 cm/s in this field. Whilst it is in the magnetic field, the dust particle is also in an electric field of strength 1.00×10^-2 V/m pointing to the north.

(a) What is the magnitude and direction of the magnetic force on the dust particle?
(b) What is the magnitude and direction of the electric force on the dust particle?
(c) What is the magnitude of the net force on the dust particle?


Homework Equations



F=|q|vB

F=(k|q1q2|) / r^2

B=E/V


The Attempt at a Solution



a) Magnitude: F=|q|vB = (-8 x 10^-18 C)(-0.003 m/s)(0.4 T) = 9.6 x 10^-21 N
Direction: Since the velocity is south and the force must be perpendicular to the velocity, the force must lie in a plane perpendicular to the north/south axis.

b) I know it has to do with "1.00 x 10^-2 V/m" but all I could do with it was:
Magnitude: B=E/V --> (1.00 x 10^-2 V/m) / (0.003 m/s) = 3.3(repeating) T
Direction: North

c) Would the net force be the sum of answers "a" and "b" or would it be:
Net force = sqrt(a^2 + b^2)
a: being the force in a)
b: being the force in b)


Thank-you.
 
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  • #2


Hi quik,

quik said:

Homework Statement



A magnet produces a 0.40 T field between its poles, directed horizontally to the east. A dust particle with charge q = -8.0×10^-18 C is moving vertically downwards with a speed of 0.30 cm/s in this field. Whilst it is in the magnetic field, the dust particle is also in an electric field of strength 1.00×10^-2 V/m pointing to the north.

(a) What is the magnitude and direction of the magnetic force on the dust particle?
(b) What is the magnitude and direction of the electric force on the dust particle?
(c) What is the magnitude of the net force on the dust particle?


Homework Equations



F=|q|vB

F=(k|q1q2|) / r^2

B=E/V


The Attempt at a Solution



a) Magnitude: F=|q|vB = (-8 x 10^-18 C)(-0.003 m/s)(0.4 T) = 9.6 x 10^-21 N
Direction: Since the velocity is south and the force must be perpendicular to the velocity, the force must lie in a plane perpendicular to the north/south axis.

Okay; now you can use the right hand rule and the charge of the particle to determine whether the force is to the north or the south. What do you get?

b) I know it has to do with "1.00 x 10^-2 V/m" but all I could do with it was:
Magnitude: B=E/V

No, I don't believe this equation applies here. You need the relationship between electric force and electric field. What equation is that?

--> (1.00 x 10^-2 V/m) / (0.003 m/s) = 3.3(repeating) T
Direction: North

c) Would the net force be the sum of answers "a" and "b" or would it be:
Net force = sqrt(a^2 + b^2)
a: being the force in a)
b: being the force in b)

If the two forces to be added are parallel or anti-parallel, you will either add or subtract the magnitudes to find the total magnitude; if the forces are perpendicular, you use the pythagorean theorem to find the magnitude.


Thank-you.
 
  • #3


Hello alphysicist, thanks for replying.

alphysicist said:
Okay; now you can use the right hand rule and the charge of the particle to determine whether the force is to the north or the south. What do you get?

I'm a little confused here, I thought:
+z axis = out of screen
+x axis = B
-y axis = v

But if the force is either North or South, since the velocity is going south, wouldn't it be north?

alphysicist said:
No, I don't believe this equation applies here. You need the relationship between electric force and electric field. What equation is that?

E = F/q --> F = Eq --> (1.00 x 10^-2 V/m)(-8 x 10^-18 C) = -8 x 10^-20 N

alphysicist said:
If the two forces to be added are parallel or anti-parallel, you will either add or subtract the magnitudes to find the total magnitude; if the forces are perpendicular, you use the pythagorean theorem to find the magnitude.

Would the net force be: (9.6 x 10^-21 N) + (-8 x 10^-20 N) = -7.04 x 10^-20 N ..?

Thanks again
 
  • #4


quik said:
Hello alphysicist, thanks for replying.



I'm a little confused here, I thought:
+z axis = out of screen
+x axis = B
-y axis = v

But if the force is either North or South, since the velocity is going south, wouldn't it be north?

This answer is correct, but it seems like you might be misunderstanding. The velocity is not to the south, it is vertically downwards. The force is perpendicular to both the velocity and the magnetic field (which narrows the choices down to two directions) and the right hand rule picks which one of those directions is correct.

E = F/q --> F = Eq --> (1.00 x 10^-2 V/m)(-8 x 10^-18 C) = -8 x 10^-20 N

That looks right (but I think you want to write it explicitly as a magnitude and direction).

Would the net force be: (9.6 x 10^-21 N) + (-8 x 10^-20 N) = -7.04 x 10^-20 N ..?

That looks right (but it is not quite in the form of a magnitude yet).

Thanks again
 
  • #5


alphysicist said:
This answer is correct, but it seems like you might be misunderstanding. The velocity is not to the south, it is vertically downwards. The force is perpendicular to both the velocity and the magnetic field (which narrows the choices down to two directions) and the right hand rule picks which one of those directions is correct.

Magnitude: 9.6 x 10^-21 N
Direction: North

alphysicist said:
That looks right (but I think you want to write it explicitly as a magnitude and direction).

Magnitude: -8 x 10^-20 N
Direction: South

alphysicist said:
That looks right (but it is not quite in the form of a magnitude yet).
[/QUOTE]

Magnitude: 7.04 x 10^-20 N
Direction: South
 
  • #6


quik said:
Magnitude: 9.6 x 10^-21 N
Direction: North



Magnitude: -8 x 10^-20 N
Direction: South




Magnitude: 7.04 x 10^-20 N
Direction: South

Magnitudes cannot be negative, but once you fix that everything looks right to me.
 

Related to Calculating the Forces on a Charged Particle in an Electric and Magnetic Field

1. How do you calculate the force on a charged particle in an electric field?

The force on a charged particle in an electric field can be calculated using the equation: F = qE, where F is the force, q is the charge of the particle, and E is the electric field strength.

2. What is the direction of the force on a charged particle in an electric field?

The direction of the force on a charged particle in an electric field is determined by the direction of the electric field itself. The force will be in the direction of the electric field if the particle is positively charged, and opposite to the direction of the electric field if the particle is negatively charged.

3. How does the presence of a magnetic field affect the force on a charged particle in an electric field?

If a charged particle is moving in an electric field and a magnetic field is also present, the force on the particle will be the sum of the forces due to the electric and magnetic fields. The direction of the magnetic force will be perpendicular to both the particle's velocity and the magnetic field.

4. What is the equation for calculating the magnetic force on a charged particle in a magnetic field?

The magnetic force on a charged particle in a magnetic field can be calculated using the equation: F = qvBsinθ, where F is the force, q is the charge of the particle, v is the velocity of the particle, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field.

5. How do you determine the net force on a charged particle in both an electric and magnetic field?

To determine the net force on a charged particle in both an electric and magnetic field, you must first calculate the individual forces on the particle due to each field. Then, you can use vector addition to find the resultant force, taking into account the direction and magnitude of each force. The net force will determine the acceleration of the particle in the combined fields.

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