There exists a uniform magnetic field of magn | vedclass

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(D) Given: $B = 1\, T$,$E = 1\, V/m$,$m = 1\, kg$,$q = 1\, C$,$\vec{v}_0 = 1\hat{i}\, m/s$. The electric field is along the $y-$ axis,so the accelerat

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$(0, \pi^2/2, 2)$

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(D) Given: $B = 1\, T$,$E = 1\, V/m$,$m = 1\, kg$,$q = 1\, C$,$\vec{v}_0 = 1\hat{i}\, m/s$. The electric field is along the $y-$ axis,so the acceleration in the $y-$ direction is $a_y = \frac{qE}{m} = \frac{1 	imes 1}{1} = 1\, m/s^2$. The position in the $y-$ direction at time $t$ is $y(t) = v_{0y}t + \frac{1}{2}a_yt^2 = 0 + \frac{1}{2}(1)t^2 = \frac{t^2}{2}$. At $t = \pi$,$y = \frac{\pi^2}{2}$.The magnetic field is along the $y-$ axis. The Lorentz force $\vec{F} = q(\vec{v} 	imes \vec{B})$ causes circular motion in the $x-z$ plane. The radius of this circular path is $R = \frac{mv_{\perp}}{qB} = \frac{1 	imes 1}{1 	imes 1} = 1\, m$. The angular frequency is $\omega = \frac{qB}{m} = 1\, rad/s$. The position in the $x-z$ plane is given by $x(t) = R \sin(\omega t)$ and $z(t) = R(1 - \cos(\omega t))$ (since it starts at origin with velocity along $x$). At $t = \pi$,$x(\pi) = 1 \sin(\pi) = 0$ and $z(\pi) = 1(1 - \cos(\pi)) = 1(1 - (-1)) = 2$. Thus,the coordinates are $(0, \pi^2/2, 2)$.

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