Question

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./s. 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 $\Omega$]}

• A. 197
• B. 100
• C. 80
• D. 280

Hint:

For a small loop deep inside a long solenoid carrying AC, only the time-varying flux contributes to EMF. $e = -A \frac{dB}{dt}$.

Answer 1

The Correct Option is A

Since the loop is 10 cm inside a 100 cm long solenoid, it is in the uniform magnetic field region. Motional EMF is zero.
The magnetic field is $B(t) = \mu_0 n I(t)$.
$\Phi = B \cdot A = (\mu_0 n I_0 \sin \omega t) (\pi r^2)$.
$\Phi = (4\pi \times 10^{-7})(500)(10 \sin \omega t)(\pi \times (10^{-2})^2)$.
$\Phi = 2\pi \times 10^{-3} \sin \omega t \times \pi \times 10^{-4} = 2\pi^2 \times 10^{-7} \sin \omega t \text{ Wb}$.
Induced EMF $e = -\frac{d\Phi}{dt} = -2\pi^2 \times 10^{-7} \omega \cos \omega t$.
With $\omega = 1000$, peak EMF $e_0 = 2\pi^2 \times 10^{-4} \text{ V}$.
Peak current $i_0 = \frac{e_0}{R} = \frac{2\pi^2 \times 10^{-4}}{10} = 2\pi^2 \times 10^{-5} \text{ A}$.
RMS current $i_{rms} = \frac{i_0}{\sqrt{2}} = \frac{2\pi^2 \times 10^{-5}}{\sqrt{2}}$.
We are given $i_{rms} = \frac{\alpha}{\sqrt{2}} \mu A = \frac{\alpha \times 10^{-6}}{\sqrt{2}}$.
Equating the terms: $\alpha \times 10^{-6} = 2\pi^2 \times 10^{-5}$.
$\alpha = 20 \pi^2 \approx 20 \times 9.87 = 197.4$.
Nearest integer is 197.