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(D) The specific charge is given by $\alpha = q/m$. The particle moves in a helical path in a uniform magnetic field.The time period of the helical mo

Vedclass · Accepted Answer

$\left( \frac{V_o \pi}{B_o \alpha}, 0, - \frac{2 V_o}{B_o \alpha} \right)$

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(D) The specific charge is given by $\alpha = q/m$. The particle moves in a helical path in a uniform magnetic field.The time period of the helical motion is $T = \frac{2 \pi m}{B_o q} = \frac{2 \pi}{B_o \alpha}$.The given time is $t = \frac{\pi}{B_o \alpha} = \frac{T}{2}$.The velocity component parallel to the magnetic field is $v_x = V_o$,which remains constant. Thus,the $x$-coordinate at $t = T/2$ is $x = v_x \cdot t = V_o \cdot \frac{\pi}{B_o \alpha} = \frac{V_o \pi}{B_o \alpha}$.The velocity component perpendicular to the magnetic field is $v_y = V_o$. The particle undergoes circular motion in the $yz$-plane with radius $r = \frac{m v_y}{B_o q} = \frac{V_o}{B_o \alpha}$.At $t = T/2$,the particle completes half a circle in the $yz$-plane. Starting from $(0, 0)$ in the $yz$-plane with initial velocity in the $+y$ direction,after half a period,the displacement in the $y$-direction is $0$ and in the $z$-direction is $-2r = -\frac{2 V_o}{B_o \alpha}$.Therefore,the coordinates are $\left( \frac{V_o \pi}{B_o \alpha}, 0, - \frac{2 V_o}{B_o \alpha} \right)$.

$A$ charged particle of specific charge $\alpha$ is released from the origin at time $t = 0$ with velocity $\vec{V} = V_o \hat{i} + V_o \hat{j}$ in a magnetic field $\vec{B} = B_o \hat{i}$. The coordinates of the particle at time $t = \frac{\pi}{B_o \alpha}$ are (specific charge $\alpha = q/m$):

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