Question

Two long coaxial solenoids of radius $R$ and $2R$ have equal number of turns per unit length. They carry time-varying currents $i_1=2kt$ and $i_2=kt$ respectively, in the same direction. A point charge released between the solenoids at a distance $r$ , is seen to move along a circular path. Then the value of $r$ is

• A. $r=R\sqrt{2}$
• B. $r=3R/2$
• C. $r=R\sqrt{3}$
• D. It can be any value between $R$ and $2R$

Answer 1

Answer: (A)

The magnetic field due to the inner solenoid (within its volume) is
$B_1={\left(\mu \right)}_{0}n\left(2kt\right)$
The magnetic field due to the outer solenoid (within its volume) is
$B_2={\left(\mu \right)}_{0}n\left(kt\right)$
The flux of magnetic field through the circular area of radius $r$ is
$\varphi =B_1\pi R^2+B_2\pi r^2$
$\Rightarrow \varphi =\pi {\left(\mu \right)}_{0}nkt\left(2R^2+r^2\right)$
If $\overrightarrow{E}$ is the induced electric field at $r$ , then
$\oint \overrightarrow{E}.d\overrightarrow{r}=-\frac{d\varphi }{dt}$
$\Rightarrow E\times 2\pi r=\pi {\left(\mu \right)}_{0}nk\left(2R^2+r^2\right)$
$\Rightarrow E={\left(\mu \right)}_{0}nk\left(\frac{R^2}{r}+\frac{r}{2}\right)$
The radius of the circle traversed by the charged particle is
$r=\frac{mv}{q{B}_2}$
$v=E\frac{qt}{m}={\left(\mu \right)}_{0}nk\left(\frac{R^2}{r}+\frac{r}{2}\right)\frac{qt}{m}$
$\Rightarrow {\left(\mu \right)}_{0}nk\left(\frac{R^2}{r}+\frac{r}{2}\right)\frac{qt}{m}=\frac{qr{B}_2}{m}$
$\Rightarrow {\left(\mu \right)}_{0}nkt\left(\frac{R^2}{r}+\frac{r}{2}\right)=r{\left(\mu \right)}_{0}nkt$
$\Rightarrow R^2+\frac{r^2}{2}=r^2$
$\Rightarrow r=R\sqrt{2}$