Answer
(a) $2.3mN$
(b) $65^{\circ}$ measured from the positive z axis toward the negative y axis in the yz plane
Work Step by Step
We can find the magnitude and direction of the required magnetic force as follows:
$F_1=qvBsin\theta$
$F_1=34\times 10^{-6}C\times 73m/s\times 0.40T sin90^{\circ}$
$F_1=9.928\times 10^{-4}N$
According to the right hand rule, its direction is along the positive z-direction.
and $F_2=qvBsin\theta$
$F_2=34\times 10^{-6}C\times 73m/s\times 0.85Tsin90^{\circ}$
$F_2=2.1907\times 10^{-3}N$ and it is directed along the negative y-direction.
Now $F=\sqrt{F_1^2+F_2^2+2F_1F_2cos\theta}$
$F=\sqrt{F_1^2+F_2^2}$ as $\theta=90^{\circ}\space and cos90^{\circ}=0$
$F=\sqrt{(9.928\times 10^{-4}N)^2+(2.1097\times 10^{-3})^2}$
$F=2.3\times 10^{-3}N=2.3mN$
and $\theta=tan^{-1}(\frac{F_2}{F_1})$
$\theta=tan^{-1}\frac{2.1097\times 10^{-3}N}{9.928\times 10^{-4}N}=65^{\circ}$
