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Alpha partides \left(m=6.7 \times 10^{-27} {~kg}\right) and q=+2e are accelerated from rest through a potential difference of 6.7 {kv}. Then, they enter a nugnetei field B=0.2T perpendicular to them direction of their motion. The radius of the path described

Option: 1

8.375 m


Option: 2

8.375 cm


Option: 3

0


Option: 4

None


Answers (1)

best_answer

r=\frac{m v}{q B}

and v=\sqrt{\frac{2 k}{m}}

so \begin{aligned} \gamma & =\frac{\sqrt{2 \mathrm{Km}}}{q B} \\ \end{aligned}

           \begin{aligned} & =\frac{\sqrt{2 \times 6.7 \times 10^3 \times 2 \times 1.6 \times 10^{-19} \times 6.7 \times 10^{-27}}}{2 \times 1.6 \times 10^{-19} \times 0.2} \\ \end{aligned}     

           \begin{aligned} =\frac{6.7}{8} \times 10^{-1} \\ \end{aligned}

           \begin{aligned} =\frac{6.7}{80} \\ \end{aligned}

            \begin{aligned} =0.08375 \mathrm{~m}=8.375 \mathrm{~cm} \end{aligned}

 

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manish

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