A long solenoid of radius 2 R contains another coaxial solenoid of half the radius . The coils have the same number of turns per unit length and initially both  carry no current. At the  same instant, the current in both  solenoids start increasing linearly with time . At any moment the  current flowing in the inner coil is twice as large as that in the outer one and their directions are  the same . Due the increasing currents, a charged particle  , initially at rest between the Solenoids, starts moving along a circular trajectory ( see Figure) of radius  $\text{r=}\sqrt{\text{K}}\text{R.}$ Then find K.

 

Let  the Current in the outer coil at any instant of time , say t, be $I=\alpha t$ and in the inner coil is $2I=2\alpha t$, where $\alpha$ is a constant. Because of these currents the magnetic field in the  outer coil is $B={\mu }_{0}n\left(\alpha t\right)$ , where  n is the number of turns per unit length . Because  in the annular part (region between two coils )  no field will be there due to the inner coil, because the  annular region is the region outside the inner coil, so ${\left({\text{B}}_{\text{annular}}\right)}_{\text{inner\hspace{0.17em}coil}}\text{=0\hspace{0.17em}\hspace{0.17em}and}{\left({\text{B}}_{\text{annular}}\right)}_{\text{outer\hspace{0.17em}coil\hspace{0.17em}}}{\text{=μ}}_{\text{0}}\text{nI}$ 

Similarly, the field in the Region II inside the inner coil will be

$\begin{array}{rl} & B_{\text{REGIONII}}=B_{\text{outer coil}}+B_{\text{imner coil}}\\ & \Rightarrow B_{\text{REGIONII}}=\mu nI+{\mu }_{0}n\left(2I\right)=3{\mu }_{0}nI=3B\end{array}$

The magnetic flux enclosed by the particle ‘s trajectory of radius   r  is

$\begin{array}{rl} & \varphi =\left(\pi R^2\right)\left(2B\right)+\left(\pi r^2\right)\left(B\right)\\ & \Rightarrow \varphi =(2R^2+r^2)\pi {\mu }_{0}n\alpha t\end{array}$

Since we have

$\begin{array}{rl} & |\text{∮}\overrightarrow{E}\cdot d\overrightarrow{\ell }|=\frac{d\varphi }{dt}\\ & \Rightarrow E\left(2\pi r\right)=\frac{d\varphi }{dt}\\ & \Rightarrow E\left(2\pi r\right)=(2R^2+r^2)\pi {\mu }_{0}n\alpha \end{array}$

$\Rightarrow E=\frac{\left(2R^2+r^2\right)}{r}\frac{{\mu }_{0}n\alpha }{2}$...........(1)

The charged particle is held in its circular orbit by the magnetic field , and so we have 

$\frac{m{\upsilon }^2}{r}=q\upsilon B$..............(2)

Further, due to the induced electric field the particle gets  accelerated along its circular orbit by the tangenttial  component of the net force.  $\Rightarrow m{a}_t=qE$ where  m is the mass and q the electric charge of the  particle. Further, since the magnitude of the electric field is constant, the speed of the particle increases uniformly with time . So, 

$\upsilon =a_{t}t=\left(\frac{qE}{m}\right)t$

Substituting the value of E from (1), we get

$\upsilon =\frac{q}{m}\left(\frac{2R^2+r^2}{r}\frac{{\mu }_{0}n\alpha }{2}\right)t$...............(3)

Again , substituting this value of $\upsilon$ and the  value of $B\left(={\mu }_{0}n\alpha t\right),$ into equation (2), We get

$\begin{array}{rl} & \frac{m}{r}\frac{(2R^2+r^2)}{r}\frac{{\mu }_{0}n\alpha }{2}\frac{q}{m}t=q{\mu }_{0}n\alpha t,\\ & \Rightarrow \frac{(2R^2+r^2)}{2r^2}=1\\ & \Rightarrow r=\sqrt{2R}\\ & \Rightarrow K=2\end{array}$