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

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

Q78JEE Main 2020 · 2 SEP · SHIFT 2Medium

Moving Charge in a Magnetic Field

The figure shows a region of length ‘l’ with a uniform magnetic field of 0.3 T in it and a proton entering the region with velocity 4 ×\times 105 ms–1 making an angle 60° with the field. If the proton completes 10 revolution by the time it cross the region shown, ‘l’ is close to

(mass of proton = 1.67 ×\times 10–27 kg, charge of the proton = 1.6 ×\times 10–19 C)

Figure

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Answer: (d)

  1. Pitch p=vcos⁡θ⋅T=vcos⁡θ⋅2πmqBp = v\cos\theta \cdot T = v\cos\theta \cdot \dfrac{2\pi m}{qB}.
  2. p=4×105×0.5×2π×1.67×10−271.6×10−19×0.3=2×105×2.19×10−7≈0.0437p = 4 \times 10^{5} \times 0.5 \times \dfrac{2\pi \times 1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 0.3} = 2 \times 10^{5} \times 2.19 \times 10^{-7} \approx 0.0437 m
  3. In 10 revolutions: l=10p≈0.44l = 10p \approx 0.44 m
  4. So the answer is (d) 0.44 m.
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