An electron moving with a speed u along the p | vedclass

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(D) The magnetic force acting on a charged particle is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$. Since the magnetic force is always perpendicula

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$v = u, y < 0$

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(D) The magnetic force acting on a charged particle is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$. Since the magnetic force is always perpendicular to the velocity of the particle,it does no work on the particle. Therefore,the kinetic energy and the speed of the electron remain constant. Thus,$v = u$.Using the right-hand rule for the magnetic force on an electron (charge $q = -e$): $\vec{F} = -e(\vec{v} 	imes \vec{B})$. Given $\vec{v} = u\hat{i}$ and $\vec{B} = -B_0\hat{k}$,we have $\vec{F} = -e(u\hat{i} 	imes -B_0\hat{k}) = -e(uB_0\hat{j}) = -euB_0\hat{j}$.Initially,the force is directed along the negative $y-$ axis. This causes the electron to follow a circular path that curves downwards. Consequently,when the electron exits the magnetic field region,its coordinate $y$ will be negative $(y < 0)$.

An electron moving with a speed $u$ along the positive $x-$ axis at $y = 0$ enters a region of uniform magnetic field $\vec{B} = -B_0\hat{k}$ which exists to the right of the $y-$ axis. The electron exits from the region after some time with the speed $v$ at coordinate $y$. Then:

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