Physics - Current Electricity Question with Solution | TestHub

Particle moves in xy plane. Its path is not a circle because B is not constant. However, speed of the particle remains constant at ${\mathrm{V}}_0$. Let velocity at point $\mathrm{P}(\mathrm{x},\mathrm{y})$ be inclined to x axis at an angle $\theta$. Force will be perpendicular to velocity, inclined at $\theta$ with y direction. ∴ ${\mathrm{ma}}_{\mathrm{y}}=\mathrm{F}\cos \theta \Rightarrow m\frac{d{V}_y}{dt}=\left(B_{0}x\right)q{V}_0\cos \theta$ $\Rightarrow m\frac{d{V}_y}{dx}\cdot \frac{dx}{dt}=B_{0}q\left(V_x\right)\left[\therefore V_0\cos \theta =V_x\right]\frac{dx}{dt}=V_x$ But $\frac{d{V}_y}{dx}=\frac{B_{0}q}{m}x\Rightarrow {\int }_0^{v_0}d{V}_y=\frac{B_{0}q}{m}{\int }_0^{x_0}xdx$ Where $x_0=$ maximum $x$ co-ordinate of the particle where velocity becomes parallel to y axis. $\therefore V_0=\frac{B_{0}q}{2m}{x}_0^2\Rightarrow x_0=\sqrt{\frac{2m{V}_0}{B_{0q}}}$, Magnetic force at $x=x_0$ is $F=q{V}_0\left(B_0{x}_0\right)$ $\therefore F=q{V}_0{B}_0\sqrt{\frac{2m{V}_0}{B_{0}q}}=v_0\sqrt{2mq{B}_0{v}_0}$ If radius of curvature of path if R then $\frac{m{v}_0^2}{R}=F\Rightarrow \frac{m{V}_0^2}{R}=v_0\sqrt{2mq{B}_0{v}_0}\Rightarrow R=\sqrt{\frac{m{V}_0}{2q{B}_0}}$