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  • A charge particle of 2μC accelerated by a potential difference of 100 V enters a region of uniform magnetic field of magnitude 4mT at right angle to the direction of field. The charge par

A charge particle of 2μC accelerated by a potential difference of 100 V enters a region of uniform
magnetic field of magnitude 4mT 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 ______ ×  10^{-18}  kg.

Option: 1

144


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & R=\frac{\mathrm{mV}}{\mathrm{qB}}=\frac{\mathrm{p}}{\mathrm{qB}} \\ & R=\frac{\sqrt{2 \mathrm{mq} \Delta \mathrm{V}}}{\mathrm{qB}} \\ & 3 \times 10^{-2}=\frac{\sqrt{2 \mathrm{~m} \times 2 \times 10^{-6} \times 10^2}}{2 \times 10^{-6} \times 4 \times 10^{-3}} \\ & 3 \times 10^{-2} \times 2 \times 10^{-6} \times 4 \times 10^{-3}=\sqrt{4 \mathrm{~m} \times 10^{-4}} \\ \end{aligned}

\begin{aligned} & 24 \times 10^{-11}=\sqrt{4 \mathrm{~m} \times 10^{-4}} \\ & \mathrm{~m}=\frac{24 \times 24 \times 10^{-22}}{4 \times 10^{-4}} \\ & \mathrm{~m}=144 \times 10^{-18} \mathrm{Kg} \end{aligned}

 

Posted by

seema garhwal

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