A long solenoid contains another coaxial solenoid (whose radius $R$ is half of its own). Their coils have the same number of turns per unit length and initially both carry no current. At the same instant currents start increasing linearly with time in both solenoids. 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. As a result of the increasing currents charged particle, initially at rest between the solenoids, starts moving along a circular trajectory (see figure) What is the radius $r$ of the circle? 

The current at time $t$ is $I=kt$ in the outer coil, and $2I=2kt$ in the inner one, where $k$ is a constant. Because of these currents the magnetic field in the outer coil is $B={\mu }_{0}nkt$, whilst in the inner one it is $3B$, where $n$ is the number of turns per unit length. The magnetic flux enclosed by the particle’s trajectory of radius $r$ is

$\Phi =\pi R^2\times 2B+\pi r^2\times B=(2R^2+r^2)\pi {\mu }_{0}nkt$

The (constant) magnitude of the induced electric field $E$ can be calculated from the rate of change of magnetic flux with time:

$E\times 2\pi r=\frac{d\Phi }{dt}=(2R^2+r^2)\pi {\mu }_{0}nk$

And so

$E=\frac{(2R^2+r^2)}{r}\frac{{\mu }_{0}nk}{2}$

The charged particle is held in its circular orbit by the magnetic field and so from the zero net radial component of the force acting on it, we obtain

$\frac{m{v}^2}{r}=qvB\dots \dots \dots \dots \left(1\right)$

The particle is accelerated along its circular orbit by the tangential component of the net force according to $m{a}_t=qE$, where $m$ is the mass and $q$ the electric charge of the particle.

As the magnitude of the electric field is constant, the speed of the particle increases uniformly with time, 

$v=a_{t}t=\frac{qE}{m}t=\frac{(2R^2+r^2)}{r}\frac{{\mu }_{0}nk}{2}\frac{q}{m}t$

Inserting this and the value of $B$ into equation $\left(1\right)$we get, 

$\frac{m}{r}\frac{(2R^2+r^2)}{r}\frac{{\mu }_{0}nk}{2}\frac{q}{m}t=q{\mu }_{0}nkt$

which is satisfied if 

$\frac{(2R^2+r^2)}{2r^2}=1$, i.e. $r=\sqrt{2}R$