Capacitance Question 6
Question 6 - 01 February - Shift 1
A charge particle of $2 \mu C$ accelerated by a potential difference of $100 V$ enters a region of uniform magnetic field of magnitude $4 mT$ at right angle to the direction of field. The charge particle completes semicircle of radius $3 cm$ inside magnetic field. The mass of the charge particle is $\times 10^{-18} kg$.
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Answer: (144)
Solution:
Formula: Lorentz-force
OR
$ r=\frac{m v}{q B}=\frac{\sqrt{2 k m}}{q B}, m=\frac{r^{2} q^{2} B^{2}}{2 k} $
$ \begin{gathered} m=\frac{\frac{1}{100} \times \frac{3}{100} \times 2 \times 2 \times 4 \times 10^{-3} \times 4 \times 10^{-3} \times 10^{-12}}{2 \times(100) \times 10^{-6}} \\ \quad=144 \times 10^{-18} kg \end{gathered} $
