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An electrons of mass 9.1 \times 10^{-31} \mathrm{~kg} under the action of fields of magnets moves in a circle of radius 3.5 \mathrm{~cm} at a speed of 3.5 \times 10^{-6} \mathrm{~ms}^{-1}. If a proton of mass 1.67 \times 10^{-27} \mathrm{~kg} were to move in a circle of the same radius in the same magnetic field find its speed?

Option: 1

1908


Option: 2

1907


Option: 3

1906


Option: 4

1905


Answers (1)

best_answer

\begin{aligned} &\text { mass of electron }=9.1 \times 10^{-31} \mathrm{~K} g\\ &\begin{gathered} r_e=3.5 \times 10^{-2} \mathrm{~m} \\ v_e=3.5 \times 10^{-6} \mathrm{~ms}^{-1} \\ m_p=1.67 \times 10^{-27} \mathrm{~kg} \\ v_p=? \quad r_p=3.5 \times 10^{-2} \mathrm{~m} \end{gathered} \end{aligned}

\begin{aligned} & q v B=m v^2 / r \Rightarrow v=q r B / m \\ & \Rightarrow \frac{v_p}{v_e}=\frac{q_p r_p B / m_p}{q_e r_e B / m_l}=\frac{m_e}{m_p} \\ & \Rightarrow V_p=V_e \times \frac{m_e}{m_p} \\ & \end{aligned}

\begin{aligned} & =\frac{3.5 \times 10^{-6} \times 9.1 \times 10^{-31}}{1.67 \times 10^{-27}} \\ & =19.0718 \times 10^{-31+27+6} \\ & =19.0718 \times 10^2=1907.18 \\ & \simeq 1907 \end{aligned}

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manish painkra

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