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This is the numerical question.

A particle having a mass of 0.05 mg carries a charge of $5\times10^{-6}C$. The particle is given an initial horizontal velocity of 5000 m/s, what is the magnitude and direction of the magnetic field that will keep the particle moving in a horizontal direction?

Well I didn't understand this question. what I think is once the charged particle enters the magnetic field whatever magnitude it is it will travel in a circular path. I don't understand why this question is asking for the magnitude of magnetic field so that it will keep moving in horizontal direction.whatever the magnitude is,it should move in a circular path is what i have read but why doesn't it do so?

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2 Answers

up vote 2 down vote accepted

I would guess the question is asking you to calculate the magnetic field that produces a force exactly equal to the gravitational force. That is, without the magnetic field the particle will accelerate downwards due to gravity. You need the field that exactly balances this to the particle carries on in a straight line.

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then it means that the magnitude of magnetic field will make it keep moving in st.line and it will not bend in circular path. –  AaKASH Apr 22 '13 at 9:24
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mass (m)= 0.05 mg = 0.05*10^-6 kg charge(q)=5*10^-5 C velocity(v)=5000 m/s for the charge move to constant horizontal direction; F=Bqvsin0 and mg=F
Also, the velocity must be perpendiculars to B. So, mg=Bqv B=0.5*10^-6*9.8/5*10^-6*5000 Therefore, B= 1.96*10^-5 T

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Although your effort is very nice, we discourage simply giving the numbers for the solution to homework problems and prefer If you'd clarify the conceptual problems the OP has. That way, the answer can be useful for people working on different problems involving the same principle. –  Neuneck Jul 31 '13 at 10:17
    
Dear Ashokkunwar, you can use Latex to improve your answer. –  Ali Jul 31 '13 at 10:17
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