Two long coaxial solenoids of radius R and 2R have equal number of...

The magnetic field due to the inner solenoid (within its volume) is

${B}_{1} = {μ}_{0}n \left(2kt\right)$

The magnetic field due to the outer solenoid (within its volume) is

${B}_{2} = {μ}_{0}n \left(kt\right)$

The flux of magnetic field through the circular area of radius $r$ is

$ϕ = {B}_{1}\piR^{2} + {B}_{2}\pir^{2}$

$⇒ ϕ = \pi{μ}_{0}nkt \left(2R^{2} + r^{2}\right)$

If $\overset{→}{E}$ is the induced electric field at $r$, then

$∮ \overset{→}{E}. d\overset{→}{r} = - \dfrac{dϕ}{dt}$

$⇒ E \times2\pir = \pi{μ}_{0}nk \left(2R^{2} + r^{2}\right)$

$⇒ E = {μ}_{0}nk \left(\dfrac{R2}{r} + \dfrac{r}{2}\right)$

The radius of the circle traversed by the charged particle is

$r = \dfrac{mv}{q{B}_{2}}$

$v = E\dfrac{qt}{m} = {μ}_{0}nk\left(\dfrac{R2}{r} + \dfrac{r}{2}\right)\dfrac{qt}{m}$

$⇒ {μ}_{0}nk\left(\dfrac{R2}{r} + \dfrac{r}{2}\right)\dfrac{qt}{m} = \dfrac{qr{B}_{2}}{m}$

$⇒ {μ}_{0}nkt\left(\dfrac{R2}{r} + \dfrac{r}{2}\right) =$ r{μ}_{0}nkt$</p><p>$⇒ R^{2} + $\dfrac{r2}{2} =$ r^{2}$</p><p>$⇒ r = R\sqrt{2}$</p>$