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 $xy$-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 $N)$ is ________.

Answer 1

Answer: 10.01

$F_2$ 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 $90^{\circ }$.
$F_2=\left(10^{-2}\right)\left(10^{-6}\right)\left(10^6\right)=10^{-2}N\approx 10.01N$