Diverting a beam of charged particles using M.F.

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SUMMARY

The discussion centers on calculating the minimum magnetic field required to divert a beam of charged particles with a kinetic energy of 1 keV and charge of 1.6 x 10-19 C. The formula used is Bmin = √(2mK.E.)/qr, where r is the distance of 1 cm. The initial calculation yielded Bmin as 311.1 T using a mass of 8 x 10-22 kg, but the correct answer of 1.414 T is derived using a mass of 16 x 10-27 kg, indicating a potential typographical error in the mass values provided in the problem statement.

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zorro
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Homework Statement


A beam of charged particles of K.E. = 1keV and q=1.6 x 10-19 C and 2 masses 8 x 10-22kg and 16 x 10-27kg come out of an accelerator tube and strike a plate at a distance of 1cm from the end of the tube, where the particles emerge perpendicularly. Find the value of the smallest magnetic field which can prevent the beam from striking the plate.


The Attempt at a Solution



Bmin = √(2mK.E.)/qr where r=1cm

On solving, I got
Bmin as √m x 1.1 x 1013
Now the min. value of magnetic field will be the one which can just divert the heavier particles.
So m = 8 x 10-22kg which gives Bmin as 311.1 T

The answer given is 1.414 T which is found by substituting m as 16 x 10-27kg
I don't understand how this is possible.
Is the answer wrong?
Please help!
 
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Your method looks good, though the question is not all that clear without a diagram. It sure looks like the masses got garbled when the question was typed. 8 x 10^-27kg and 16 x 10^-27kg would make a lot more sense and lead to the given answer.
 
Yes you are right. Thanks!
 

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