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(B) The magnetic force on a moving charge is given by $\vec F = q(\vec v 	imes \vec B)$. Since the electron has a negative charge $(q = -e)$,the forc

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negative $z-$ axis,negative $x-$ axis and zero

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(B) The magnetic force on a moving charge is given by $\vec F = q(\vec v 	imes \vec B)$. Since the electron has a negative charge $(q = -e)$,the force is $\vec F = -e(\vec v 	imes \vec B) = e(\vec B 	imes \vec v)$.Case $1$: $\vec v$ is along the negative $x-$ axis $(-\hat i)$ and $\vec B$ is along the negative $y-$ axis $(-\hat j)$.$\vec F = -e((-\hat i) 	imes (-\hat j)) = -e(\hat k) = -e\hat k$. This is along the negative $z-$ axis.Case $2$: $\vec v$ is along the negative $y-$ axis $(-\hat j)$ and $\vec B$ is along the positive $x-$ axis $(\hat i)$.$\vec F = -e((-\hat j) 	imes \hat i) = -e(\hat k) = -e\hat k$. Wait,let's re-evaluate: $\vec v = -v\hat j$,$\vec B = B\hat i$. $\vec v 	imes \vec B = (-v\hat j) 	imes (B\hat i) = -vB(\hat j 	imes \hat i) = -vB(-\hat k) = vB\hat k$. Since $q = -e$,$\vec F = -e(vB\hat k) = -evB\hat k$. This is along the negative $z-$ axis.Re-evaluating Case $2$ from image: $\vec v$ is along negative $y-$ axis $(-\hat j)$,$\vec B$ is along positive $x-$ axis $(\hat i)$. $\vec F = -e((-\hat j) 	imes \hat i) = -e(\hat k) = -e\hat k$. Actually,$\vec v 	imes \vec B = -\hat j 	imes \hat i = \hat k$. So $\vec F = -e(\hat k) = -e\hat k$. The force is along negative $z-$ axis.Case $3$: $\vec v$ is along negative $y-$ axis $(-\hat j)$ and $\vec B$ is along positive $y-$ axis $(\hat j)$.Since $\vec v$ and $\vec B$ are anti-parallel,$\vec v 	imes \vec B = 0$. Thus,the force is zero.

The figure shows three situations when an electron with velocity $\vec v$ travels 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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