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Protons and deuterons are accelerated through a same potential difference before they enter a region of uniform magnetic field applied perpendicular to their motion. If protons describe a circle of radius 10 cm then what will be the radius of circle of a deuteron? |
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Answer» Solution :Let us ASSUME charge q is accelerated through a potential difference of V then we can write the following: `qV = 1//2 mv^(2)` `rArrmv = sqrt(2qmV)` Radius of circular path is given by ` r = (mv)/(qB) = sqrt(2qmV)/(qB) = sqrt((2mV)/(qB^(2)))` In this case potential difference VAND magnetic field intensity B are same.. Hence, the ratio of radii of deuteron to that of proton can be WRITTEN as follows: `r_(d)/r_(p) = sqrt((2m_(d)V)/(q_(d)B^(2)))/sqrt((2m_(p)V)/(q_(p)B^(2))) = sqrt(m_(d)/m_(p) xx q_(p)/q_(d)) = sqrt((2M)/m xx e/e) = sqrt2` We know that the radius of circle of proton is 10 cm. Hence, radius of the circular path followed by deuteron would be `10 root()()2 cm` . |
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