91Ó°ÊÓ

(a) The magnetic field imposes a magnetic force on the moving charge, which is dependent on the velocity vector, as demonstrated by Ampere's experiment. The force increases as the angle between the velocity and the magnetic field increases.

$\stackrel{→}{F}=q\stackrel{→}{v}×\stackrel{→}{B}$

Solving the equation,

$\stackrel{→}{B}=\frac{\stackrel{→}{F}}{q\stackrel{→}{v}}$

The magnetic force is downward, so its direction is $-\hat{j}$while the velocity direction is to the right $\widehat{+i}.$Dividing the component $\hat{j}\text{by}\hat{i}\text{result in}\widehat{+}\hat{k}$.

$\frac{\widehat{-j}}{\hat{i}}=\widehat{+}k$

$\stackrel{→}{B}=\frac{\left(F\right)\widehat{−j}}{\left(qv\right)\widehat{i}}=\left(\frac{F}{qv}\right)\widehat{k}$

The direction is in the positive of z, so the direction of the deflection is out of the page.

(b) The magnetic force is into the page (negative direction of $z$), so its direction is ${}^{Ë„}-k$while the velocity direction is to the right ${}^{Ë„}+i$. Dividing the component ${}^{Ë„}-k$by $\hat{i}$result in${}^{Ë„}-j$.

role="math" localid="1650107281377" $\frac{{}^{Ë„}-k}{\hat{i}}={}^{Ë„}-j$

$\stackrel{→}{B}=\frac{\left(F\right)-\hat{k}}{(-qv)\hat{i}}=\left(\frac{F}{qv}\right)+\hat{j}$

The direction is in the positive of $y$, so the direction of the deflection is upward.