A charge particle of \(2\,\mu\,C\) accelerated by a potential difference of \(100\,V\) enters a region of uniform magnetic field of magnitude \(4\,m\,T\) 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 & Solution

Correct Answer

(B) \(144\)

Step-by-step Solution

Detailed explanation

\(=\frac{m v}{q B}=\frac{\sqrt{2 k m}}{q B}, m=\frac{r^2 q^2 B^2}{2 k}\)

\(m = \frac{\frac{1}{100} \times \frac{3}{100} \times 2 \times 2 \times 4 \times 10^{-3} \times 4 \times 10^{-3} \times 10^{-12}}{2 \times(100) \times 10^{-6}}\)

\(=144 \times 10^{-18}\,kg\)

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