Suppose a long solenoid of 100 cm length, radius 2 cm having 500 turns per unit length, carries a current $$I= 10 \sin (\omega t)$$ A, where $$\omega$$ = 1000 rad.ls. A circular conducting loop (B) of radius 1 cm coaxially slided through the solenoid at a speed $$v = 1 cm/s$$. The r.m.s. current through the loop when the coil B is inserted 10 cm inside the solenoid is $$\alpha/\sqrt{2}\mu A$$. The value of $$\alpha$$ is ______.
[Resistance of the loop= 10$$\Omega$$]

Step 1: Magnetic field inside the long solenoid

The magnetic field $$B$$ produced by a long solenoid is:

$$ B = \mu_0 n I $$

Given $$I = I_0 \sin(\omega t)$$, we substitute this into the equation:

$$ B = \mu_0 n I_0 \sin(\omega t) $$

Step 2: Magnetic flux through the circular loop

The flux $$\Phi$$ linked with the inner coil (B) of radius $$r_B$$is:

$$ \Phi = B \cdot A = B \cdot (\pi r_B^2) $$

$$ \Phi = \mu_0 n I_0 \pi r_B^2 \sin(\omega t) $$

Note: The sliding speed $$v$$ does not induce motional EMF because the field is uniform deep inside the solenoid, meaning $$\frac{dB}{dx} = 0$$.

Step 3: Induced EMF in the loop

By Faraday's Law, the magnitude of the induced EMF ($$\varepsilon$$) is:

$$ \varepsilon = \left| \frac{d\Phi}{dt} \right| = \mu_0 n I_0 \pi r_B^2 \omega \cos(\omega t) $$

The peak EMF ($$\varepsilon_0$$) is the amplitude of this expression:

$$ \varepsilon_0 = \mu_0 n I_0 \pi r_B^2 \omega $$

Step 4: Calculate Peak Current

The peak current $$i_0$$ is the peak EMF divided by the resistance $$R$$:

$$ i_0 = \frac{\varepsilon_0}{R} = \frac{\mu_0 n I_0 \pi r_B^2 \omega}{R} $$

$$ i_0 = \frac{(4\pi \times 10^{-7}) \times 500 \times 10 \times \pi (10^{-2})^2 \times 1000}{10} $$

$$ i_0 = \frac{4\pi \times 10^{-7} \times 5000 \times \pi \times 10^{-4} \times 1000}{10} $$

$$ i_0 = \frac{20\pi^2 \times 10^{-4}}{10} $$

$$ i_0 = 2\pi^2 \times 10^{-5} \text{ A} $$

To convert this to microamperes ($$\mu\text{A}$$), multiply by $$10^6$$:

$$ i_0 = 20\pi^2 \ \mu\text{A} $$

\Step 5: RMS Current

The root-mean-square current is related to the peak current by $$i_{rms} = \frac{i_0}{\sqrt{2}}$$:

$$ i_{rms} = \frac{20\pi^2}{\sqrt{2}} \ \mu\text{A} $$

$$ i_{rms}=\frac{197}{\sqrt{2}} \ \mu\text{A} $$