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February 7th, 2011, 01:31 AM   #1
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A proton moves perpendicular to a uniform magnetic fieldB at

A proton moves perpendicular to a uniform magnetic field B at 2.20 X 10^7 m/s and experiences an acceleration of 2.00 X 10^13 m/s2 in the +x direction when its velocity is in the +z direction.

1 ) Determine the magnitude and direction of the field.

2) drwa the case

------------------------

I try

The force here is the maximum due the charged particle moves perpendicular to the field
the magnitude of the field
Therefore, F ? qvB sin ?
B = F/qvsinQ =

(Is correct ?)

Now direction B out


2)

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