JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Magnetic Effects of Current

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A cyclotron is used to accelerate protons. If the operating magnetic field is 1.0 T and the radius of the cyclotron 'dees' is 60 cm, the kinetic energy of the accelerated protons in MeV will be:
[use

m_{p} = 1.6 \times 1 0^{- 27} kg, e = 1.6 \times 1 0^{- 19} C

]

(A)12
(B)18✓ Correct
(C)16
(D)32

Answer: (B)

Solution

The kinetic energy of a charged particle accelerated in a cyclotron is given by

K = \frac{q ^{2} B ^{2} r _{0}^{2}}{2 m}

. Substituting the given values:

q = 1.6 \times 1 0^{- 19}

B = 1.0

r_{0} = 0.6

m, and

m_{p} = 1.6 \times 1 0^{- 27}

kg, we get

K = \frac{( 1.6 \times 1 0 ^{- 19} ) ^{2} ( 1.0 ) ^{2} ( 0.6 ) ^{2}}{2 ( 1.6 \times 1 0 ^{- 27} )} = 18

MeV.

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Single correct option

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