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JEE Main 2021 (26 Aug, Shift 2) · Moving Charge in a Magnetic Field

Moving Charge in a Magnetic Field question from JEE Main 2021 (26 Aug, Shift 2)

Q72JEE Main 2021 · 26 AUG · SHIFT 2Difficult

Moving Charge in a Magnetic Field

If the maximum value of accelerating potential provided by a radio frequency oscillator is 12 kV. The number of revolutions made by a proton in a cyclotron to achieve one sixth of the speed of light is ...............

[mp = 1.67 ×\times 10−-27 kg, e = 1.6 ×\times 10−-19 C, Speed of light = 3 ×\times 108 m/s]

Numerical answer type. Work it out, then open the answer.

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Answer: 543

  1. Final KE: v=c6=5×107 m/sv = \dfrac{c}{6} = 5 \times 10^{7}\text{ m/s}, K=12(1.67×10−27)(5×107)2≈2.09×10−12 JK = \tfrac{1}{2}(1.67 \times 10^{-27})(5 \times 10^{7})^2 \approx 2.09 \times 10^{-12}\text{ J}.
  2. Energy gained per revolution (two gap crossings): 2eV=2×1.6×10−19×12×103=3.84×10−15 J2eV = 2 \times 1.6 \times 10^{-19} \times 12 \times 10^{3} = 3.84 \times 10^{-15}\text{ J}.
  3. N=2.09×10−123.84×10−15≈543N = \dfrac{2.09 \times 10^{-12}}{3.84 \times 10^{-15}} \approx 543
  4. So the number of revolutions is 543.
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