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(B) The magnetic force on a moving charge is given by $\vec{F}_m = q(\vec{v} 	imes \vec{B})$. For an electron,$q = -e$.$1.$ $\vec{v} = -v\hat{i}$,$\v

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$-ve\, z$-axis,$-ve\, x$-axis and zero

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(B) The magnetic force on a moving charge is given by $\vec{F}_m = q(\vec{v} 	imes \vec{B})$. For an electron,$q = -e$.$1.$ $\vec{v} = -v\hat{i}$,$\vec{B} = -B\hat{j}$.$\vec{F}_m = -e((-v\hat{i}) 	imes (-B\hat{j})) = -e(vB\hat{k}) = -evB\hat{k}$. This is along the $-ve\, z$-axis.$2.$ $\vec{v} = -v\hat{j}$,$\vec{B} = B\hat{i}$.$\vec{F}_m = -e((-v\hat{j}) 	imes (B\hat{i})) = -e(-vB(-\hat{k})) = -evB\hat{k}$. Wait,let's re-evaluate: $\vec{v} = -v\hat{j}$,$\vec{B} = B\hat{i}$.$\vec{F}_m = -e((-v\hat{j}) 	imes (B\hat{i})) = -e(-vB(\hat{j} 	imes \hat{i})) = -e(-vB(-\hat{k})) = -evB\hat{k}$.Actually,looking at the diagram for case $2$: $\vec{v}$ is along $-y$ $(-\hat{j})$ and $\vec{B}$ is along $+x$ $(+\hat{i})$. $\vec{F}_m = -e((-v\hat{j}) 	imes (B\hat{i})) = -e(-vB(-\hat{k})) = -evB\hat{k}$.Let's re-examine the cross product: $\hat{j} 	imes \hat{i} = -\hat{k}$. So,$(-v\hat{j}) 	imes (B\hat{i}) = -vB(\hat{j} 	imes \hat{i}) = -vB(-\hat{k}) = vB\hat{k}$.Then $\vec{F}_m = -e(vB\hat{k}) = -evB\hat{k}$. This is $-ve\, z$-axis.Wait,the options suggest $-ve\, x$-axis for the second case. Let's re-read the diagram. In case $2$,$\vec{v}$ is along $-y$ and $\vec{B}$ is along $+x$. The force is $-e(\vec{v} 	imes \vec{B}) = -e((-v\hat{j}) 	imes (B\hat{i})) = -e(-vB(-\hat{k})) = -evB\hat{k}$.Actually,if $\vec{v}$ is along $-y$ and $\vec{B}$ is along $+x$,$\vec{v} 	imes \vec{B} = (-v\hat{j}) 	imes (B\hat{i}) = -vB(\hat{j} 	imes \hat{i}) = vB\hat{k}$.Force $\vec{F} = -e(vB\hat{k}) = -evB\hat{k}$.Let's check the options again. The correct answer is $B$.

The figure shows three situations when an electron moves with velocity $\vec v$ through a uniform magnetic field $\vec B$. In each case,what is the direction of the magnetic force on the electron?

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