CBSE Class 12 · 2005 · All India · Magnetic Effects of CurrentLong Answer

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Explain the principle and working of a cyclotron with the help of a labelled diagram. A cyclotron's oscillator frequency is

10 MHz

. What should be the operating magnetic field for accelerating protons? If the radius of its dees is

60 cm

, what is the kinetic energy of the proton beam produced? Express your answer in MeV.

(e = 1.6 \times 1 0^{- 19} C, m_{p} = 1.67 \times 1 0^{- 27} kg, 1 MeV = 1.602 \times 1 0^{- 13} J .)

Answer: $B \approx 0.66 T$ and the kinetic energy $\approx 7 MeV$ .

Solution

\textbf{Principle/working:} crossed electric and magnetic fields. The magnetic field makes the proton circle inside the dees; each gap-crossing the alternating field (at the cyclotron frequency

f = \frac{q B}{2 π m}

) accelerates it, so it spirals out to larger radius with greater energy. Operating field:

B = \frac{2 π m f}{q} = \frac{2 π ( 1.67 \times 1 0 ^{- 27} ) ( 10 \times 1 0 ^{6} )}{1.6 \times 1 0 ^{- 19}} \approx 0.66 T .

Maximum speed

v = \frac{q B R}{m}

, so the kinetic energy

K = \frac{1}{2} m v^{2} = \frac{q ^{2} B ^{2} R ^{2}}{2 m} = \frac{( 1.6 \times 1 0 ^{- 19} ) ^{2} ( 0.66 ) ^{2} ( 0.60 ) ^{2}}{2 ( 1.67 \times 1 0 ^{- 27} )} \approx 1.2 \times 1 0^{- 12} J \approx 7 MeV .

CBSE Class 12 Board · 2005 · All India · Subjective

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CBSE Class 12 · 2005 · All India · Magnetic Effects of CurrentLong Answer

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