# Charged particle in B field Watch

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https://imgur.com/a/aN8li5N

Don't know on how to work this out.

Don't really understand the situation.

What is the direction of motion, the direction of the field and the direction of deflection? I don't really get it :/

How do I relate this to the angle of deflection?

I know, F = Bqv = mv^2/r

so r = mv/Bq

but no idea how to find the speed

Don't know on how to work this out.

Don't really understand the situation.

What is the direction of motion, the direction of the field and the direction of deflection? I don't really get it :/

How do I relate this to the angle of deflection?

I know, F = Bqv = mv^2/r

so r = mv/Bq

but no idea how to find the speed

Last edited by BobbJo; 1 year ago

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#2

(Original post by

https://imgur.com/a/aN8li5N

Don't know on how to work this out.

Don't really understand the situation.

What is the direction of motion, the direction of the field and the direction of deflection? I don't really get it :/

How do I relate this to the angle of deflection?

I know, F = Bqv = mv^2/r

so r = mv/Bq

but no idea how to find the speed

**BobbJo**)https://imgur.com/a/aN8li5N

Don't know on how to work this out.

Don't really understand the situation.

What is the direction of motion, the direction of the field and the direction of deflection? I don't really get it :/

How do I relate this to the angle of deflection?

I know, F = Bqv = mv^2/r

so r = mv/Bq

but no idea how to find the speed

You need a bit of geometry to do the problem. The total deflection of 20° mean how much it is deviated from the original path of motion.

Relate the radius

*R*of the arc of circular motion due to the magnetic force with the radius of the cylinder (r = 5 cm) with the total deflection of 20°.

Then use what you have set up to solve for v:

Bqv = mv^2/R

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(Original post by

You need a bit of geometry to do the problem. The total deflection of 20° mean how much it is deviated from the original path of motion.

Relate the radius

Then use what you have set up to solve for v:

Bqv = mv^2/R

**Eimmanuel**)You need a bit of geometry to do the problem. The total deflection of 20° mean how much it is deviated from the original path of motion.

Relate the radius

*R*of the arc of circular motion due to the magnetic force with the radius of the cylinder (r = 5 cm) with the total deflection of 20°.Then use what you have set up to solve for v:

Bqv = mv^2/R

The situation looks like this ?

How do I calculate what the radius of the motion is?

I don't know how to relate the radii of the circle to the cylinder and the total deflection :/

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#4

(Original post by

....

How do I calculate what the radius of the motion is?

I don't know how to relate the radii of the circle to the cylinder and the total deflection :/

**BobbJo**)....

How do I calculate what the radius of the motion is?

I don't know how to relate the radii of the circle to the cylinder and the total deflection :/

Draw a line from the point of entry to the point of exit to form a chord in a circle (this circle is the region of the magnetic field).

Draw a radius from the point of entry to the center of the circle and then draw a radius from the center of the circle to the point of exit.

In this way, an isosceles triangle is formed. The angles in this triangle is related to the angle of the total deflection.

How does the radii of the arc of circular motion extend from the point of entry and exit respectively outside the magnetic field region to the center of the circular motion?

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(Original post by

Yes but the region of magnetic field is circular NOT square or rectangle.

Draw a line from the point of entry to the point of exit to form a chord in a circle (this circle is the region of the magnetic field).

Draw a radius from the point of entry to the center of the circle and then draw a radius from the center of the circle to the point of exit.

In this way, an isosceles triangle is formed. The angles in this triangle is related to the angle of the total deflection.

How does the radii of the arc of circular motion extend from the point of entry and exit respectively outside the magnetic field region to the center of the circular motion?

**Eimmanuel**)Yes but the region of magnetic field is circular NOT square or rectangle.

Draw a line from the point of entry to the point of exit to form a chord in a circle (this circle is the region of the magnetic field).

Draw a radius from the point of entry to the center of the circle and then draw a radius from the center of the circle to the point of exit.

In this way, an isosceles triangle is formed. The angles in this triangle is related to the angle of the total deflection.

How does the radii of the arc of circular motion extend from the point of entry and exit respectively outside the magnetic field region to the center of the circular motion?

I omitted the path of the electrons in the field but A and B are points where the electrons enter and leave

I have made an attempt but don't know if this is right? The lines from the point of exit and point of entry intersect and there is an angle 20 degrees between them?

Not sure what the next step is

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#6

(Original post by

https://imgur.com/a/HQqYaIV

I omitted the path of the electrons in the field but A and B are points where the electrons enter and leave

I have made an attempt but don't know if this is right? ….

**BobbJo**)https://imgur.com/a/HQqYaIV

I omitted the path of the electrons in the field but A and B are points where the electrons enter and leave

I have made an attempt but don't know if this is right? ….

(Original post by

…

The lines from the point of exit and point of entry intersect and there is an angle 20 degrees between them? ….

**BobbJo**)…

The lines from the point of exit and point of entry intersect and there is an angle 20 degrees between them? ….

(Original post by

…..Not sure what the next step is

**BobbJo**)…..Not sure what the next step is

The red curve line is the path of travel of electron in the magnetic field. The orange line is the radius of the circular motion.

E

_{1}and E

_{2}are the point of entry and exit, respectively.

O is the centre of the magnetic field.

One of the key points is knowing how the radius of the circular motion make with the path of the circular motion.

Once this known, relate the radius

*R*of the arc of circular motion due to the magnetic force with the radius of the cylinder (

*r*= 5 cm) with the total deflection of 20°.

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(Original post by

One of the key points is knowing how the radius of the circular motion make with the path of the circular motion.

Once this known, relate the radius

**Eimmanuel**)One of the key points is knowing how the radius of the circular motion make with the path of the circular motion.

Once this known, relate the radius

*R*of the arc of circular motion due to the magnetic force with the radius of the cylinder (*r*= 5 cm) with the total deflection of 20°.
Spoiler:

Angle E1OE2 = 160 degrees

I can then work out E1E2 = 10 sin 80 = 9.85 cm

I then work out R = (5 sin 80 / sin 10) cm = 28.4 cm = 0.284m

I work out v = Bqr/m = 4.98 x 10^7 ms-1

Which is close but not the answer of 5.1 x 10^7 ms-1... I used e = 1.6 x 10^-19 C, m = 9.11 x 10^-31 kg

Working backwards from the given answer, the radius turns out to be 29.0cm. I have no idea where I went wrong...

Show

Angle E1OE2 = 160 degrees

I can then work out E1E2 = 10 sin 80 = 9.85 cm

I then work out R = (5 sin 80 / sin 10) cm = 28.4 cm = 0.284m

I work out v = Bqr/m = 4.98 x 10^7 ms-1

Which is close but not the answer of 5.1 x 10^7 ms-1... I used e = 1.6 x 10^-19 C, m = 9.11 x 10^-31 kg

Working backwards from the given answer, the radius turns out to be 29.0cm. I have no idea where I went wrong...

Edit: I just worked it out again! I got the answer. Please disregard the above.

Thanks a million for your invaluable help and insight! The value of e/m_e was given in the question as 1.8 x 10^11, which caused the difference. I truly appreciate your time! Thanks a lot!!

P.S The reason we can deduce the line extended from the exit point backwards intersects at the centre of the circle is because it enters along a radius. Is this right?

Last edited by BobbJo; 1 year ago

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#8

(Original post by

Thank you so much, I understand this a lot better now. I worked out v but I got it wrong.

Edit: I just worked it out again! I got the answer. Please disregard the above.

Thanks a million for your invaluable help and insight! The value of e/m_e was given in the question as 1.8 x 10^11, which caused the difference. I truly appreciate your time! Thanks a lot!!

**BobbJo**)Thank you so much, I understand this a lot better now. I worked out v but I got it wrong.

Spoiler:

Angle E1OE2 = 160 degrees

I can then work out E1E2 = 10 sin 80 = 9.85 cm

I then work out R = (5 sin 80 / sin 10) cm = 28.4 cm = 0.284m

I work out v = Bqr/m = 4.98 x 10^7 ms-1

Which is close but not the answer of 5.1 x 10^7 ms-1... I used e = 1.6 x 10^-19 C, m = 9.11 x 10^-31 kg

Working backwards from the given answer, the radius turns out to be 29.0cm. I have no idea where I went wrong...

Show

Angle E1OE2 = 160 degrees

I can then work out E1E2 = 10 sin 80 = 9.85 cm

I then work out R = (5 sin 80 / sin 10) cm = 28.4 cm = 0.284m

I work out v = Bqr/m = 4.98 x 10^7 ms-1

Which is close but not the answer of 5.1 x 10^7 ms-1... I used e = 1.6 x 10^-19 C, m = 9.11 x 10^-31 kg

Working backwards from the given answer, the radius turns out to be 29.0cm. I have no idea where I went wrong...

Edit: I just worked it out again! I got the answer. Please disregard the above.

Thanks a million for your invaluable help and insight! The value of e/m_e was given in the question as 1.8 x 10^11, which caused the difference. I truly appreciate your time! Thanks a lot!!

(Original post by

...P.S The reason we can deduce the line extended from the exit point backwards intersects at the centre of the circle is because it enters along a radius. Is this right?

**BobbJo**)...P.S The reason we can deduce the line extended from the exit point backwards intersects at the centre of the circle is because it enters along a radius. Is this right?

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(Original post by

Partially.

**Eimmanuel**)Partially.

Edit: I'm not sure how I justify that the tangent at the point of exit and the tangent at the point of entry will intersect at a point which is the midpoint of their x-coordinates.

I think this is not a (fully) correct explanation

Last edited by BobbJo; 1 year ago

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#10

(Original post by

What is the full reason? Is it because the path is circular? Then the tangent at the point of exit and the tangent at the point of entry will intersect at a point which is the midpoint of their x-coordinates. Then, since the tangent at the point of entry is horizontal, they must intersect at the same horizontal level as the tangent point of entry, that is along the radius. Then the point of intersection is the centre. Is this correct?

Edit: I'm not sure how I justify that the tangent at the point of exit and the tangent at the point of entry will intersect at a point which is the midpoint of their x-coordinates.

I think this is not a (fully) correct explanation

**BobbJo**)What is the full reason? Is it because the path is circular? Then the tangent at the point of exit and the tangent at the point of entry will intersect at a point which is the midpoint of their x-coordinates. Then, since the tangent at the point of entry is horizontal, they must intersect at the same horizontal level as the tangent point of entry, that is along the radius. Then the point of intersection is the centre. Is this correct?

Edit: I'm not sure how I justify that the tangent at the point of exit and the tangent at the point of entry will intersect at a point which is the midpoint of their x-coordinates.

I think this is not a (fully) correct explanation

First, we are told that "The electron beam enters the field along a radius of the cylinder", this is important and help to simplify the problem.

Next, the magnetic force is providing for the centripetal force that caused the electron travel in an arc of circle. So the electron would also exit along the radius of the cylinder.

Last edited by Eimmanuel; 1 year ago

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(Original post by

You seem to get all points but you are stitching them in a logical way.

First, we are told that "The electron beam enters the field along a radius of the cylinder", this is important and help to simplify the problem.

Next, the magnetic force is providing for the centripetal force that caused the electron travel in an arc of circle. So the electron would also exit along the radius of the cylinder.

**Eimmanuel**)You seem to get all points but you are stitching them in a logical way.

First, we are told that "The electron beam enters the field along a radius of the cylinder", this is important and help to simplify the problem.

Next, the magnetic force is providing for the centripetal force that caused the electron travel in an arc of circle. So the electron would also exit along the radius of the cylinder.

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