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A particle of mass m and charge q is projected into a region having a perpendicular magnetic field B. Find the angle of deviation of the particle as it comes out of the magnetic field if the width d of the region is slightly less than. (a) (mv)/(qB) "" (b)(mv)/(2qB) "" (c) (2mv)/(qB) |
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Answer» Solution :The RADIUS of PATH ` r=(mv)/(QB)` For `d LT r,` we have sin `theta = d/r` `(a) d= ((mv)/(qB)) = r, therefore sin theta= (((mv)/(qB)))/(((mv)/(qB)))=1 or theta = (pi)/(2) ` rad `(b) d= ((mv)/(2qB)) lt r , sin theta = d/r = ((mv)/(2qB))/((mv)/(qB)) = 1/2 or theta =pi/6 rad` `(c) d = ((2mv)/(qB))gt r`, the deviation of particle is therefore `theta= pi` rad. |
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