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JEE Main 2019 (11 Jan, Shift 1) · Moving Charge in a Magnetic Field

Moving Charge in a Magnetic Field question from JEE Main 2019 (11 Jan, Shift 1)

Q89JEE Main 2019 · 11 JAN · SHIFT 1Medium

Moving Charge in a Magnetic Field

In an experiment, electrons are accelerated, from rest, by applying a voltage of 500 V. Calculate the radius of the path if a magnetic field 100 mT is then applied. [Charge of the electron = 1.6 ×\times 10–19 C Mass of the electron = 9.1 ×\times 10–31 kg]

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Answer: (a)

  1. r=1B2mVer = \dfrac{1}{B}\sqrt{\dfrac{2mV}{e}}
  2. 2mVe=2×9.1×10−31×5001.6×10−19=5.69×10−9\dfrac{2mV}{e} = \dfrac{2 \times 9.1 \times 10^{-31} \times 500}{1.6 \times 10^{-19}} = 5.69 \times 10^{-9}, square root =7.54×10−5= 7.54 \times 10^{-5}
  3. r=7.54×10−50.1≈7.5×10−4r = \dfrac{7.54 \times 10^{-5}}{0.1} \approx 7.5 \times 10^{-4} m
  4. So the answer is (a) 7.5×10−47.5 \times 10^{-4} m.
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