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?

Explanation

Let us assume charge q is accelerated through a potential difference of V then we can write the following:
`qV = 1//2 mv^(2)`
`rArr mv = 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` .