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JEE Main 2020 (2 Sep, Shift 1) · Moving Charge in a Magnetic Field

Moving Charge in a Magnetic Field question from JEE Main 2020 (2 Sep, Shift 1)

Q79JEE Main 2020 · 2 SEP · SHIFT 1Medium

Moving Charge in a Magnetic Field

A beam of protons with speed 4 × 105 ms–1 enters a uniform magnetic field of 0.3 T at an angle of 60° to the magnetic field. The pitch of the resulting helical path of protons is close to: (Mass of the proton = 1.67 ×\times 10–27 kg, charge

of the proton = 1.69 ×\times 10–19 C)

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Answer: (d)

  1. Pitch p=vcos⁡60∘⋅2πmqBp = v\cos 60^\circ \cdot \dfrac{2\pi m}{qB}.
  2. p=2×105×2π×1.67×10−271.69×10−19×0.3=2×105×2.07×10−7≈0.041p = 2 \times 10^{5} \times \dfrac{2\pi \times 1.67 \times 10^{-27}}{1.69 \times 10^{-19} \times 0.3} = 2 \times 10^{5} \times 2.07 \times 10^{-7} \approx 0.041 m
  3. So the answer is (d) 4 cm.
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