Physics - Magnetism Question with Solution | TestHub

$\alpha =\frac{q}{m}$

$\mathrm{r}=\frac{{\mathrm{mv}}_0}{\mathrm{qB}}=\frac{{\mathrm{v}}_0}{\alpha {\mathrm{B}}_0}[$ since $\frac{\mathrm{q}}{\mathrm{m}}=\alpha ]$

T

Along the $x$-axis: $v_x=v\cos \theta (\theta =\omega t)$

Along the $y$-axis: $v_y=v\sin \theta (\theta =\omega t)$

So, $v_x=v_0\cos \omega t$ and $v_y=v_0\sin \omega t$. The angular velocity is $\omega =\frac{2\pi }{T}$.

$T=\frac{2\pi m}{qB}\Rightarrow \omega =\frac{2\pi }{2\pi m/qB}=\frac{qB}{m}=\alpha B\Rightarrow \omega =\alpha B$

Therefore, $v_x=v_0\cos (\alpha Bt)$ and $v_y=v_0\sin (\alpha Bt)$.