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JEE Main 2023 (1 Feb, Shift 1) · Moving Charge in a Magnetic Field

Moving Charge in a Magnetic Field question from JEE Main 2023 (1 Feb, Shift 1)

Q47JEE Main 2023 · 1 FEB · SHIFT 1Medium

Moving Charge in a Magnetic Field

A charge particle of 2 μC2\ \mu\text{C} accelerated by a potential difference of 100 V100\text{ V} enters a region of uniform magnetic field of magnitude 4 mT4\text{ mT} at right angle to the direction of field. The charge particle completes semicircle of radius 3 cm3\text{ cm} inside magnetic field. The mass of the charge particle is __________ ×10−18 kg\times 10^{-18}\text{ kg}

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

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Answer: 144

  1. From qV=p22mqV = \dfrac{p^2}{2m} and p=qBrp = qBr: m=qB2r22Vm = \dfrac{qB^2r^2}{2V}.
  2. m=2×10−6×(4×10−3)2×(3×10−2)22×100m = \dfrac{2 \times 10^{-6} \times (4 \times 10^{-3})^2 \times (3 \times 10^{-2})^2}{2 \times 100}
  3. m=2×16×9×10−16200=1.44×10−16m = \dfrac{2 \times 16 \times 9 \times 10^{-16}}{200} = 1.44 \times 10^{-16} kg
  4. So the answer is 144×10−18144 \times 10^{-18} kg, i.e. 144.
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