Magnetic Induction

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Homework Statement


A beam of electrons accelerated by a p.d. of 100V flies into a homogeneous magnetic field applied perpendicular to the plane of the paper towards the observer. The width of the field OP=L1=2cm. In the absence of m.f. the electron beam produces a spot at a point F on fluorescent screen AK, which is at a distance L2=6cm from teh edge of the m.f. When the m.f. is switched on the spot moves onto A along the path OQA. Find the displacement FA of the spot if the induction of the magnetic field B = 1.5 x 10-3 T


The Attempt at a Solution



Without explaining the concept- I am writing what I did in short.
Velocity gained by the electron, v= √(2eV/m)
v = 5.93 x 106m/s
r = 0.0225m

Now for the motion of the electrons along the path OQ,
Sx = uxt
Sx = 0.02m
so t = 3.373 x 10-9 s
Sy = 0.5 x v2t2/r
Sy = 8.889 x 10-3m

tanθ = Sy/Sx
θ=23.96 degrees.
This will be the angle AQE as well
so tanθ = AE/QE
from here AE = 0.0267m

So total displacement AF = AE + Sy = 0.0267 + 8.889 x 10-3
AF = 3.56cm

The answer given is 12.87cm
 

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Answers and Replies

  • #2
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If your answer is wrong, maybe you should state your concepts.
From what I can see, you calculated it as though your electrons are deflected in an eletric field.
Also i do not see, what r is.
 
  • #3
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r is the radius of the circular path traced inside the region of the magnetic field.
I used equations of motion taking acceleration as v2/r. What made you think I calculated it like that of an electric field region?
 
  • #4
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You are mixing a lot of formula from different areas together.
Acceleration v^2/r is the radial acceleration on a circular orbit. You cannot use it in this problem other than to calculate the radius. I don't know how you calculated the radius otherwise.
I thought you calculated as if in an electric field, because you calculated the time of flight through the field as Sx/ux. With ux constant. This is only correct if the force is acting always in the same direction, in this coordinate system the y direction. For a magnetic field this is not the case. The Force is always perpendicular to the velocity, so the velocity in x direction will change too.

I see two ways you can calculate this. Either by using the circle, or by calculation the actual time dependent speed in x and y direction and from that the time needed to traverse the field.
 
  • #5
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I got r from r=mv/qB (v is already calculated)
I used the same equation (Sx=uxt and Sy=0.5ayt2) for another question involving magnetic field and got an exact answer.
Here it is -

A proton of velocity 107 m/s is projected at right angles to a unifrom magnetic field of induction 100mT. How much is the particle path deflected from a straight line after it has traversed a distance of 1cm?

Sy = 0.5 x ay x t2
here ay = v2/r
where r=mv/qB
Also t = 0.01/v
Substituting all these we get Sy as 4.848 x 10-5

How do you explain this?
 
  • #6
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This will be just a coincidence. The equation you use is valid only for a constant force in y direction which is not the case here. The acceleration you get for y is only the initial acceleration in that direction. It will decrease later.
 
  • #7
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Then how do you solve such kind of problems the correct way?
 
  • #8
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The general way is to solve the complete equations of motion for the particle in the magnetic field, that is, solve the differential equations for the x and y direction. In the magnetic field these two equations will be coupled, so you have to do some tricks to solve it.
In this case there is a faster way. As you know, the particle will move on a circle in the magnetic field. The starting point is O, the endpoint is Q. As you know the radius and the x coordinate of Q you can determine they value. This gives you the deflection in the magnetic field. After that, as you correctly stated, the particle will continue to go on in the last direction it had, i.e. on a tangent to the circle. Draw the circle and you will find the correct angle. Mind that it is not the angle you calculated above.
See how far you get with that, and ask if something is unclear.
 
  • #9
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Everything is clear!! Thanks alot.
 
  • #10
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Well done.
 

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