Two long coaxial solenoids of radius R and 2 R have equal number of turns per unit length. They carry time-varying currents i 1 = 2 k t and i 2 = k t respectively, in the same direction. A point charg

Question

Two long coaxial solenoids of radius R and 2 R have equal number of turns per unit length. They carry time-varying currents i 1 = 2 k t and i 2 = k t 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

TestHub · Accepted Answer

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

$B_{1} = μ_{0} n (2 k t)$

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

$B_{2} = μ_{0} n (k t)$

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

$ϕ = B_{1} π R^{2} + B_{2} π r^{2}$

$\Rightarrow ϕ = π μ_{0} n k t (2 R^{2} + r^{2})$

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

$\oint \vec{E} . d \vec{r} = - \frac{d ϕ}{d t}$

$\Rightarrow E \times 2 π r = π μ_{0} n k (2 R^{2} + r^{2})$

$\Rightarrow E = μ_{0} n k (\frac{R^{2}}{r} + \frac{r}{2})$

The radius of the circle traversed by the charged particle is

$r = \frac{m v}{q B_{2}}$

$v = E \frac{q t}{m} = μ_{0} n k (\frac{R^{2}}{r} + \frac{r}{2}) \frac{q t}{m}$

$\Rightarrow μ_{0} n k (\frac{R^{2}}{r} + \frac{r}{2}) \frac{q t}{m} = \frac{q r B_{2}}{m}$

$\Rightarrow μ_{0} n k t (\frac{R^{2}}{r} + \frac{r}{2}) = r μ_{0} n k t$

$\Rightarrow R^{2} + \frac{r^{2}}{2} = r^{2}$

$\Rightarrow r = R \sqrt{2}$

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