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JEE Main 2022 (27 Jul, Shift 2) · Moving Charge in a Magnetic Field

Moving Charge in a Magnetic Field question from JEE Main 2022 (27 Jul, Shift 2)

Q60JEE Main 2022 · 27 JUL · SHIFT 2Medium

Moving Charge in a Magnetic Field

A cyclotron is used to accelerate protons. If the operating magnetic field is 1.0 T1.0\text{ T} and the radius of the cyclotron 'dees' is 60 cm60\text{ cm}, the kinetic energy of the accelerated protons in MeV will be: [use mp=1.6×10−27 kg,e=1.6×10−19 C[\text{use } m_p = 1.6 \times 10^{-27}\text{ kg}, e = 1.6 \times 10^{-19}\text{ C} ]

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Answer: (b)

  1. Maximum kinetic energy in a cyclotron: K=q2B2R22mK = \dfrac{q^2B^2R^2}{2m}.
  2. K=(1.6×10−19)2(1.0)2(0.6)22×1.6×10−27=2.88×10−12 JK = \dfrac{(1.6 \times 10^{-19})^2 (1.0)^2 (0.6)^2}{2 \times 1.6 \times 10^{-27}} = 2.88 \times 10^{-12}\text{ J}
  3. In MeV: 2.88×10−121.6×10−13=18 MeV\dfrac{2.88 \times 10^{-12}}{1.6 \times 10^{-13}} = 18\text{ MeV}
  4. So the answer is (b) 18.
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