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# HELP Electromagtism Physics Please watch

1. I also get 2.5cm, I think it's probably safe to assume the answer given is a misprint.

Just to check, for the horizontal speed I got 1.87 x 10 ^ 7 and for the vertical I got 9.39 x 10 ^5

I did the distance from slit just by ratio & proportion based on these.
2. (Original post by phys981)
I also get 2.5cm, I think it's probably safe to assume the answer given is a misprint.

Just to check, for the horizontal speed I got 1.87 x 10 ^ 7 and for the vertical I got 9.39 x 10 ^5

I did the distance from slit just by ratio & proportion based on these.
The first one is correct but for the second isn't it to power of 6 ^6. (answer says 9.4*10^6)
3. In that case 25cm would be correct.
I'll check back to see what I did wrong....
4. (Original post by phys981)
In that case 25cm would be correct.
I'll check back to see what I did wrong....
Oh damn... what did I do wrong then, because i had those numbers but I still didn't manage to get 25cm
5. (Original post by MrToodles4)
Oh damn... what did I do wrong then, because i had those numbers but I still didn't manage to get 25cm
OK, what I did wrong was to change 10cm to 0.01m instead of 0.1m of which I am deeply ashamed....

So now for the last bit,

well jsut as you drew a triangle with the vertical and horizontal speeds, we can do another one with the distances. The horizontal distance is 50cm, the vertical is what we want to find.

So i said

9.4 ^ 6 / 1.87 ^7 = distance from silt / 0.5

and that should now give the right answer

Alternatively you could use trig with the angle you worked out before.
6. (Original post by phys981)
OK, what I did wrong was to change 10cm to 0.01m instead of 0.1m of which I am deeply ashamed....

So now for the last bit,

well jsut as you drew a triangle with the vertical and horizontal speeds, we can do another one with the distances. The horizontal distance is 50cm, the vertical is what we want to find.

So i said

9.4 ^ 6 / 1.87 ^7 = distance from silt / 0.5

and that should now give the right answer

Alternatively you could use trig with the angle you worked out before.

Ah don't worry man lool happens to everyone.

But those ratios... damn how can i even visualise that... I drew the triangles but Im having a hard time understanding the reasoning behind those ratios

But yeah I managed to do it with the trig
7. similar triangles - glad you got it anyway.

8. (Original post by phys981)
similar triangles - glad you got it anyway.

That makes perfect sense now You are actually amazing. I don't know how to thank you. Tbh I think I pestered you enough - part g and h are still bothering me but I may have to try it another day idk. Like I have no idea what g is even entailing

and with part h Im not sure which equation. Would I use left hand rule? so Wouldn't the conventional current go form right to left? and so how is the direction of magnetic field act INTO the page like the answer says...
9. For g I think

If you increase the anode-cathode voltage the electrons would be travelling faster so would spend less time between the plates - what would that mean for their final vertical speed (you can assume here that the electric field stays the same) - and hence for the deflection?

If on the other hand you increase the voltage or field between the plates, you increase the vertical acceleration - what happens to the deflection? (assume here that the horizontal speed hasn't been changed.

Then for h you're back to the balance between electric and magnetic forces, so like the other question you had
10. (Original post by phys981)
For g I think

If you increase the anode-cathode voltage the electrons would be travelling faster so would spend less time between the plates - what would that mean for their final vertical speed (you can assume here that the electric field stays the same) - and hence for the deflection?

If on the other hand you increase the voltage or field between the plates, you increase the vertical acceleration - what happens to the deflection? (assume here that the horizontal speed hasn't been changed.

Then for h you're back to the balance between electric and magnetic forces, so like the other question you had
for the first instance for g; vertical speed would be less? so less deflection.

For increased voltage between the places you increase vertical accelerate and so deflection increases! Ah I get it now. But how comes in the first instance you talk about speed - "travelling faster" - is this still acceleration? What Im trying to ask is basically if you increase voltage between anode cathode you get faster movement due to higher acceleration?

And for h I used F=BILsinx and obviously i had to work out current by doing rate of charge. And I managed to get the correct answer but Im not sure about the direction - why would it be inside the plane - I used the left hand rule.. Its a negative charge going from left to right so conventionally it should be right to left? How do I know which way the force acts?
11. For the first instance for g, correct. And yes to higher acceleration BUT in this first instance remember that the acceleration comes BEFORE the electron is fired up towards the plates, so once it leaves that anode, it's travelling at constant speed - but faster than it was before. (the other way to think about it, the way it was originally asked in teh question, is that the KE it gains in teh cathode-anode bit is higher)

Then for h, yes, I'd go for Bqv but what you have is the same. For the direction, well this time we have electrons so the electric field is going to deflect them upwards. We need the magnetic field to deflect them downwards so use the left hand rule to see which direction the B field should be. Yes, the conventional current in this bit is right to left.
12. (Original post by phys981)
For the first instance for g, correct. And yes to higher acceleration BUT in this first instance remember that the acceleration comes BEFORE the electron is fired up towards the plates, so once it leaves that anode, it's travelling at constant speed - but faster than it was before. (the other way to think about it, the way it was originally asked in teh question, is that the KE it gains in teh cathode-anode bit is higher)

Then for h, yes, I'd go for Bqv but what you have is the same. For the direction, well this time we have electrons so the electric field is going to deflect them upwards. We need the magnetic field to deflect them downwards so use the left hand rule to see which direction the B field should be. Yes, the conventional current in this bit is right to left.
Ohhh so in that first instance, acceleration is only in the cathode anode bit so it gets faster but when it leaves the anode its speed is constant so therefore acceleration would now be constant/o.

And the rest makes sense, so we just need a force to bring the electrons back down to their horizontal lines and theres using LHR we get a magnetic field inside the plane.

FINALLY this exercise is now complete. Thank you so much for everything, I shall continue with this hardcore topic and more questions on it tommoro
13. No problem.

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