A charged particle carrying charge $q=1\mu C$ moves in uniform magnetic fields with velocity $v_1={10}^6 m/s$ at angle $45°$ with x-axis in the $xy$-plane and experience a force $F_1=5\sqrt{2} mN$ along the negative z-axis. When the same particle moves with velocity $v_2={10}^{6}m/s$ along the z-axis, it experience a force $F_2$ in y-direction. Find

(a) the magnitude and direction of the magnetic field

(b) the magnitude of the force $F_2$

$F_2$ is in direction when velocity is along z-axis. Therefore, magnetic field should be along x-axis. So let, $B=B_0\hat{i}$ 

(a) Given,      $v_1=\frac{{10}^6}{\sqrt{2}}\hat{i}+\frac{{10}^6}{\sqrt{2}}\hat{j}$

and $F_1=-5\sqrt{2}\times {10}^{-3}\hat{k}$

From the equation, $F=q(v\times B)$

We have,

 $$\begin{gathered} (-5\sqrt{2}\times {10}^{-3})\hat{k}=\left[\left(\frac{{10}^6}{\sqrt{2}}\hat{i}+\frac{{10}^6}{\sqrt{2}}\hat{j}\right)\times \left(B_0\hat{i}\right)\right] \\ =\frac{B_0}{\sqrt{2}}\hat{k} \\ \therefore \frac{B_0}{\sqrt{2}}=5\sqrt{2}\times {10}^{-3} \\ {B}_0={10}^{-2 }T \end{gathered}$$

Therefore, the magnetic field is $B=\left({10}^{-2 }\hat{i}\right) T$

(b) $F_2=B_{0}q{v}_2\sin 90°$

As the angle between B and v in this case is $90°.$

$\therefore F_2=\left({10}^{-2}\right)\left({10}^{-6}\right)\left({10}^6\right)$

$={10}^{-2}N$