Physics - Magnetism Question with Solution | TestHub

$$\begin{array}{rl}& \vec{L}=L\cos \theta \hat{i}-L\sin \theta \hat{j}\\ & {\vec{F}}_{+}=q(\vec{V}+\vec{\omega }\times \vec{L}/2)\times \vec{B}\\ & {\vec{F}}_{-}=q(-\vec{v}+\vec{\omega }\times \vec{L}/2)\times \vec{B}\\ & {\vec{F}}_{+}+{\vec{F}}_{-}=q(\vec{\omega }\times \vec{L})\times \vec{B}\\ & =q[(\omega B)\vec{L}-(\vec{L}\cdot \vec{B})\vec{\omega }]=q\omega B\vec{L}\\ & {\vec{F}}_{+}-{\vec{F}}_{-}=2q(\vec{V}\times \vec{B})\\ & {\vec{F}}_{net}=2m\frac{d\vec{v}}{dt}={\vec{F}}_{+}+{\vec{F}}_{-}\\ & \frac{d\vec{v}}{dt}=\frac{Bq\omega }{2m}\vec{L}\dots (1)\\ & \frac{d\vec{v}}{dt}=\frac{Bq}{2m}\frac{d\theta }{dt}\\ & \frac{d\vec{V}}{dt}=\frac{Bq}{2m}\vec{L}d\theta \\ & \int d\vec{v}={\int }_0^{\theta }\frac{BqL}{2m}(\cos \theta d\theta \hat{i}-\sin \theta d\theta \hat{j})\\ & (\because \vec{L}=L\cos \theta \hat{i}-L\sin \theta \hat{j})\\ & \vec{v}=\frac{BqL}{2m}[\sin \theta \hat{i}-(1-\cos \theta )\hat{j}]\\ & {\vec{\tau }}_{net}=\frac{\vec{L}}{2}\times {\vec{F}}_{+}(2)\\ & \frac{m{L}^2}{2}\frac{d\vec{\omega }}{dt}=\frac{\vec{L}}{2}\times (2q(\vec{v}\times \vec{B}))\\ & =q(\vec{B}\cdot \vec{L})\vec{v}-q(\vec{L}\cdot \vec{v})\vec{B}\\ & \frac{m{L}^2}{2}\frac{d\vec{\omega }}{dt}=-q(\vec{L}\cdot \vec{v})\vec{B}\\ & \frac{d\vec{\omega }}{dt}=-\frac{2q(\vec{L}+{\vec{F}}_{-}}{m{L}^2})\\ & \Rightarrow \frac{d\omega }{dt}=\frac{-2qB}{m{L}^2}(\vec{L}\cdot \vec{V})\dots (3)\\ & \vec{V}\cdot \vec{L}=\frac{Bq{L}^2}{2m}(\sin \theta \cdot \cos \theta +\sin \theta (1-\cos \theta ))\\ & \vec{V}\cdot \vec{L}=\frac{Bq{L}^2}{2m}\sin \theta \\ & \frac{d\omega }{dt}=\frac{-2Bq}{m{L}^2}\cdot \frac{Bq{L}^2}{2m}\sin \theta (from(3))\end{array}$$

$$\begin{array}{rl}& \omega \frac{d\omega }{d\theta }=\frac{-B^2{q}^2}{m^2}\sin \theta \\ & {\int }_{{\omega }_0}\omega d\omega =-\frac{B^2{q}^2}{m^2}{\int }_0^{\theta }\sin \theta \cdot d\theta \\ & \frac{{\omega }^2-{\omega }_0^2}{2}=-\frac{B^2{q}^2}{m^2}(1-\cos \theta )\\ & {\omega }^2={\omega }_0^2-\frac{2B^2{q}^2}{m^2}(1-\cos \theta )\\ & \Rightarrow {\omega }_0^2⩾\frac{4B^2{q}^2}{m^2}\\ & {\omega }_0⩾\frac{2Bq}{m}\Rightarrow {\omega }_c=\frac{2Bq}{m}\end{array}$$

This is the critical angular velocity, below which the rod will not complete one turn. When ${\omega }_0={\omega }_{\mathrm{c}}$,

$$\begin{array}{rl}& \Rightarrow {\omega }^2=\frac{4B^2{q}^2}{m^2}-\frac{2B^2{q}^2}{m^2}(1-\cos \theta )\\ & \Rightarrow \omega =\frac{2Bq}{m}\cos (\theta /2)\end{array}$$By the Work-Energy Theorem:

$$\begin{array}{rl}& {\mathrm{W}}_{\mathrm{B}}={\mathrm{KE}}_{\mathrm{f}}-{\mathrm{KE}}_{\mathrm{i}}\\ & 0=\frac{1}{2}\times \left(\frac{{\mathrm{mL}}^2}{2}\right){\omega }^2+\frac{1}{2}\cdot 2{\mathrm{mv}}^2-\frac{1}{2}\left(\frac{{\mathrm{mL}}^2}{2}\right){\omega }_0^2\\ & \Rightarrow {\omega }^2={\omega }_0^2-\frac{4}{{\mathrm{L}}^2}{\mathrm{v}}^2\\ & \Rightarrow {\mathrm{V}}^2=\frac{{\mathrm{B}}^2{\mathrm{q}}^2{\mathrm{L}}^2}{2{\mathrm{m}}^2}(1-\cos \theta )\\ & \Rightarrow \mathrm{V}=\frac{\mathrm{BqL}}{\mathrm{m}}\sin (\theta /2)\\ & \Rightarrow {\mathrm{V}}_{\max }=\frac{\mathrm{BqL}}{\mathrm{m}}=\frac{{\omega }_{\mathrm{c}}\mathrm{L}}{2}\text{ at }\theta =\pi \end{array}$$

Now, let ${\omega }_0={\omega }_{\mathrm{c}}$.

$$\begin{array}{rl}& \Rightarrow \omega =\frac{2Bq}{m}\cos (\theta /2)\\ & \frac{dx}{dt}=V_x=\frac{BqL}{2m}\sin \theta \\ & {\int }_0^{x}dx=\frac{BqL}{2m}{\int }_0^{\theta }\sin \theta \cdot dt\\ & x=\frac{BqL}{2m}{\int }_0^{\theta }\frac{\sin \theta }{\omega }d\theta \\ & x=\frac{BqL}{2m}{\int }_0^{\theta }\frac{\sin \theta \cdot d\theta }{\frac{2Bq}{m}\cdot \cos (\theta /2)}\\ & =\frac{L}{2}{\int }_0^{\theta }\sin (\theta /2)d\theta \\ & x=L\left(1-\cos \left(\frac{\theta }{2}\right)\right)\\ & \Rightarrow x_{\max }=2L\text{ when }\theta =2\pi \end{array}$$