Question

A charged particle carrying charge $q=1\, \mu C$ moves in uniform magnetic field with velocity $v_{1}=10^{6} \,m / s$ at angle $45^{\circ}$ with $x$-axis in the $x y$-plane and experiences 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 experiences a force $F_{2}$ in $y$-direction. The magnitude of the force $F_{2}$ (in $\left.N \right)$ is ________.

Answer 1

$F_{2}$ is in $y$-direction when velocity is along $z$-axis. Therefore, magnetic field should be along $x$-axis. So, let
$B=B_{0} \hat{i}$
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 equation $F=q(v \times B)$
We have
$\left(-5 \sqrt{2} \times 10^{-3}\right) \hat{k}=\left(10^{-6}\right)\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} $
$\therefore \frac{B_{0}}{\sqrt{2}}=5 \sqrt{2} \times 10^{-3}$
or $ B_{0}=10^{-2} T$
Therefore, the magnetic field is $B=\left(10^{-2} \hat{i}\right) T$
$F_{2}=B_{0} q v_{2} \sin 90^{\circ}$
As the angle between $B$ and $v$ in this case is $90^{\circ}$.
$F_{2}=\left(10^{-2}\right)\left(10^{-6}\right)\left(10^{6}\right)=10^{-2} N \approx 10.01 \,N$