If the velocity of an electron is (2 hat{i} + | vedclass

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(D) The magnetic force on a moving charge is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.Given $\vec{v} = (2 \hat{i} + 3 \hat{j}) 	ext{ m s}^{-1}$

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the direction of its velocity will change.

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(D) The magnetic force on a moving charge is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.Given $\vec{v} = (2 \hat{i} + 3 \hat{j}) 	ext{ m s}^{-1}$ and $\vec{B} = 4 \hat{k} 	ext{ T}$.Since $\vec{v}$ lies in the $xy$-plane and $\vec{B}$ is along the $z$-axis,the velocity vector is perpendicular to the magnetic field $(\vec{v} \perp \vec{B})$.The magnetic force $\vec{F}$ acts perpendicular to both $\vec{v}$ and $\vec{B}$.Because the magnetic force is always perpendicular to the velocity,it does no work on the electron,meaning the speed (kinetic energy) remains constant.However,the force causes a change in the direction of the velocity,resulting in circular motion.Therefore,the direction of its velocity will change.

If the velocity of an electron is $(2 \hat{i} + 3 \hat{j}) \text{ m s}^{-1}$ and it enters a magnetic field of $4 \hat{k} \text{ T}$,then . . . . . . .

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