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

[quik]

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.

 

[alphysicist]

Hi quik,

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.

 

[quik]

Hello alphysicist, thanks for replying.

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?

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

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

 

[alphysicist]

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

 

[quik]

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

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

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

 

[alphysicist]

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.