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(B) The $\alpha-$ particle moves in a region with $\vec{E} = E_0 \hat{i}$ and $\vec{B} = B_0 \hat{i}$. The electric force $\vec{F}_E = q_{\alpha} E_0

Vedclass · Accepted Answer

$10^7 \ m/s$

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(B) The $\alpha-$ particle moves in a region with $\vec{E} = E_0 \hat{i}$ and $\vec{B} = B_0 \hat{i}$. The electric force $\vec{F}_E = q_{\alpha} E_0 \hat{i}$ acts along the $x-$axis,causing acceleration $a_x = \frac{q_{\alpha} E_0}{m_{\alpha}}$. The magnetic force $\vec{F}_B = q_{\alpha} (\vec{v} 	imes \vec{B})$ is perpendicular to the velocity and the magnetic field,so it does no work and does not change the speed.At time $t$,the velocity component along the $x-$axis is $v_x = a_x t = \left( \frac{q_{\alpha} E_0}{m_{\alpha}} ight) \left( \sqrt{3} 	imes 10^7 \frac{m_{\alpha}}{q_{\alpha} E_0} ight) = \sqrt{3} 	imes 10^7 \ m/s$.The velocity component along the $y-$axis remains constant at $v_y = v_0$.The final speed $v_f$ is given as $2v_0$. Thus,$v_f^2 = v_x^2 + v_y^2 = (2v_0)^2 = 4v_0^2$.Substituting the values: $(\sqrt{3} 	imes 10^7)^2 + v_0^2 = 4v_0^2$.$3 	imes 10^{14} = 3v_0^2 \implies v_0^2 = 10^{14} \implies v_0 = 10^7 \ m/s$.

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