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JEE Main 2022 (30 Jun, Shift 1) · Moving Charge in a Magnetic Field

Moving Charge in a Magnetic Field question from JEE Main 2022 (30 Jun, Shift 1)

Q61JEE Main 2022 · 30 JUN · SHIFT 1Medium

Moving Charge in a Magnetic Field

A cyclotron is working at a frequency of 10 MHz. If the radius of its dees is 60 cm. The maximum kinetic energy of accelerated proton will be:

(Take: e = 1.6 ×\times 10−-19 C, mp = 1.67 ×\times 10−-27 kg)

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Answer: (a)

  1. Kmax=12mv2K_{max} = \tfrac{1}{2}mv^2 with v=2πfRv = 2\pi fR, so Kmax=2π2mf2R2K_{max} = 2\pi^2 m f^2 R^2.
  2. Kmax=2(9.87)(1.67×10−27)(107)2(0.6)2≈1.19×10−12 JK_{max} = 2(9.87)(1.67 \times 10^{-27})(10^{7})^2(0.6)^2 \approx 1.19 \times 10^{-12}\text{ J}
  3. In MeV: 1.19×10−121.6×10−13≈7.4 MeV\dfrac{1.19 \times 10^{-12}}{1.6 \times 10^{-13}} \approx 7.4\text{ MeV}
  4. So the answer is (a) 7.4 MeV.
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