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JEE Main 2022 (28 Jun, Shift 1) · Moving Charge in a Magnetic Field

Moving Charge in a Magnetic Field question from JEE Main 2022 (28 Jun, Shift 1)

Q58JEE Main 2022 · 28 JUN · SHIFT 1Medium

Moving Charge in a Magnetic Field

A singly ionized magnesium atom (A = 24) ion is accelerated to kinetic energy 5 keV, and is projected perpendicularly into a magnetic field B of the magnitude 0.5 T. The radius of path formed will be _____________ cm.

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

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Answer: 10

  1. r=2mKqBr = \dfrac{\sqrt{2mK}}{qB} with m=24×1.67×10−27≈4×10−26m = 24 \times 1.67 \times 10^{-27} \approx 4 \times 10^{-26} kg, K=5×103×1.6×10−19=8×10−16K = 5 \times 10^{3} \times 1.6 \times 10^{-19} = 8 \times 10^{-16} J.
  2. 2mK=2×4×10−26×8×10−16=64×10−42=8×10−21\sqrt{2mK} = \sqrt{2 \times 4 \times 10^{-26} \times 8 \times 10^{-16}} = \sqrt{64 \times 10^{-42}} = 8 \times 10^{-21}
  3. r=8×10−211.6×10−19×0.5=0.1r = \dfrac{8 \times 10^{-21}}{1.6 \times 10^{-19} \times 0.5} = 0.1 m
  4. So the radius is 10 cm.
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