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Q.

The radius of the dees of a cyclotron is 60cm and the magnetic field applied is 650 mT. The maximum KE of a proton accelerated through it is
$[m_{p} = 1 .67 \times 10^{- 27} kg, q = 1 .6 \times 10^{- 19} c]$

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a

$12 \times 10^{13} J$

b

$6 \times 10^{- 19} J$

c

$1 .2 \times 10^{10} J$

d

$0.6 \times 10^{- 10} J$

answer is A.

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Detailed Solution

$\begin{array}{rcl} qvB & = & \frac{{mv}^{2}}{r} \\ m v & = & q B r \\ K E & = & \frac{{(m v)}^{2}}{2 m} = \frac{{(q B r)}^{2}}{2 m} \\ = & \frac{{(1.6 \times 10^{- 19} \times 650 \times 10^{- 3} \times 60 \times 10^{- 2})}^{2}}{2 \times 1.67 \times 10^{- 27}} \\ = & 11.65 \times 10^{- 13} \end{array}$

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The radius of the dees of a cyclotron is 60cm and the magnetic field applied is 650 mT. The maximum KE of a proton accelerated through it is mp=1.67×10−27kg, q=1.6×10−19c