Magnetic Field Problem...please Help.

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INTRODUCTION:This is a problem from "Concepts Of Physics" by H.C.Verma.The chapter is "Magnetic Field".

EXACT PROLEM: "An electron is projected horizontally with a kinetic energy of 10 keV.A magnetic field of strength 10^-7 T exists in the vertically upward direction. Calculate the sideways deflection of the electron in travelling through 1m. Make appropriate approximations."

PROBLEM FACED: 1>Travelling through 1m? which 1m? 2> how to calculate the deflection?

WHAT MY BRAIN SUGGESTED: I calculated the radius using √(2Km)/qB. The particale, since it is negatively charged, will be deflected towards its left, the magnetic force acting towards its right.We have to find the deviation after it has travelled 1m, but how?

CONCLUSION: I am really having problems understanding the above problem. i have just learned "magnetic field", so my concepts are still weak. Please help if you can. Thanks a lot...

maverick280857

First of all, please draw a diagram if you haven't already done so.

The situation is as follows: the electron has some KE to start with, so it tends to go straight but as soon as it enters the region which has a magnetic field, it gets deflected and deviates from a straight line path.

First you should be able to figure out the direction of this deviation. You are right about the direction, but if you have any doubts, read the following paragraph:

Assume that the initial velocity of the electron is along the +x axis, the magnetic field is along the +y axis. As the electron is negatively charged, it gets deflected in the -z direction. If you are uncomfortable with this, use [itex]\vec{F}=-e(\vec{v}\times\vec{B})[/itex] with [itex]\vec{v} = v\hat{i}[/itex] and [itex]\vec{B} = B\hat{j}[/itex]. Thsi gives [itex]\vec{F} = -evB\hat{k}[/itex].

As the electron travels horizontally, it also gets deflected by the magnetic field...so you have to find out the sideways deflection when the horizontal distance traveled is 1m.

What is the radius of the path? Its is simply [itex]R = \frac{mv}{qB}[/itex] which can be written as [itex]R = \frac{\sqrt{2mK}}{qB}[/itex] as you say. Now the question is...is [itex]R = 1m[/itex]? If so, the electron completes a quarter circle and its sideways deflection is merely equal to 1m. Can you see why? (I haven't plugged in your values so I don't know). Even if you get the answer, it would be instructive to figure out how to handle the cases R>1m and R<1m. Draw a top down diagram (as if you are looking down along the negative y axis) of the electron's circular trajectory and you'll see what I'm saying here.

I recommend reading Resnick/Halliday vol 2 or Resnick/Hallidy/Krane vol 2. And do draw diagrams for even the most trivial seeming questions. And feel free to gets your doubts clarified here. Just keep posting your solutions (in accordance with PF rules).

Hope that helps...

thiotimoline

Does the question state that the path of the deflected electron is a circle? If yes, can use equation of UCM? If not?

maverick280857

What makes you think that the path is not circular? And why would the question mention it? Its one of the things you have to figure out.

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thanks for helping...

nrqed

The 1 meter seems to be the actual distance travelled by the electron (if I read the question properly). So that would be the *arc* of the circle followed by the electron. You also have the radius, so you should be able to figure out the angle (recall, for theta given in radians, you have [itex] s = R \theta [/itex] where "s" is the arc of the circle.

Now it becomes a geometry problem. Draw an arc of circle with the proper angle. You should be able to construct a right angle triangle with the hypothenuse being 1 meter, one side being what you are looking for and the angle being known.

Patrick

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Hey i'm not getting it that way. The answer should be around 1.5 cm but i dont know how. I'm getting the angle theta as 1.186E-13. But after that i can't really follow. Please explain. And is the 1m travelled along the arc or is it 1m along the horizontal direction? If you have time please let me know about both cases. This problem is important for our exam. Please help. Thanks a lot coz i only have words...

maverick280857

Try what nrqed said. But maybe 1m is the lateral (sweep) distance...

Why don't you post your (complete) working and then have it discussed/checked here.