JEE MAIN - Physics (2023 - 1st February Morning Shift - No. 21)
A charge particle of $$2 ~\mu \mathrm{C}$$ accelerated by a potential difference of $$100 \mathrm{~V}$$ enters a region of uniform magnetic field of magnitude $$4 ~\mathrm{mT}$$ at right angle to the direction of field. The charge particle completes semicircle of radius $$3 \mathrm{~cm}$$ inside magnetic field. The mass of the charge particle is __________ $$\times 10^{-18} \mathrm{~kg}$$
Answer
144
Explanation
$ r=\frac{m v}{q B}=\frac{\sqrt{2 k m}}{q B}, $
$m=\frac{r^2 q^2 B^2}{2 k}$
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$$ \begin{aligned} \mathrm{m}= & \frac{\frac{1}{100} \times \frac{3}{100} \times 2 \times 2 \times 4 \times 10^{-3} \times 4 \times 10^{-3} \times 10^{-12}}{2 \times(100) \times 10^{-6}} \\\\ & =144 \times 10^{-18} \mathrm{~kg} \end{aligned} $$
$m=\frac{r^2 q^2 B^2}{2 k}$
$$ \begin{aligned} \mathrm{m}= & \frac{\frac{1}{100} \times \frac{3}{100} \times 2 \times 2 \times 4 \times 10^{-3} \times 4 \times 10^{-3} \times 10^{-12}}{2 \times(100) \times 10^{-6}} \\\\ & =144 \times 10^{-18} \mathrm{~kg} \end{aligned} $$
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