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Magnetism

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25 Dec 2007 23:39:45 IST

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An electron is projected horizontally with a kinetic energy of 10Kev.A magnetic field of strength 1*10^-7 .calculate the sideways deflectionof electron travelling through 1m

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Niteesh Mehra

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25 Dec 2007 23:48:42 IST

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the question is not clear dude..Wat u can do is equate 10x10^3=1/2 mv^2 or r=sq.rt(2mKE)/qB

Bipin Dubey

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5 Jan 2008 01:25:54 IST

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Please rewrite the question. Its not clear.

Karthik M

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5 Jan 2008 01:33:16 IST

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heres how you do it:

KE of the electron is given, so use the equation KE = 1/2mv²

and get v. This will be the horizontal component of the velocity of the electron, which will remain unchanged during the course of its motion, as the magnetic field is perpendicular (should be, though its not mentioned) to the motion of the electron.

Now, the magnetic field will exert a force of Bqv on the electron which will produce an acceleration which has the magnitude Bqv/m.

Then use kinematics equation, ie, s = ut + 1/2at²

Here u=0.

So, S = 1/2at²

To get t : it is simply 1m divided by the horizontal component of velocity calculated in the first part.

Note mass of electron = 9.31 x 10^-31kgs

Substitute the values, get the answer. Get back if you have doubts

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