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

To solve this problem, we need to determine the r.m.s. (root mean square) current through a circular loop as it is moved inside a solenoid. Let's break it down step-by-step: 

1. Calculate the magnetic field inside the solenoid using the formula for the magnetic field in a solenoid: \(B = \mu_0 n I\), where \(\mu_0\) is the permeability of free space, \(n\) is the number of turns per unit length, and \(I\) is the current. Given that \(n = 500\)
2. Note the nature of the magnetic field: The magnetic field inside the solenoid is not constant as the current varies sinusoidally with time. We will express this field as: \(B(t) = \mu_0 \times 500 \times 10\sin(1000t)\).
3. Determine the EMF induced in the loop: According to Faraday's law of electromagnetic induction, the induced emf (\(\varepsilon\)) in a loop is given by: \(\varepsilon = - \frac{d\Phi}{dt}\), where \(\Phi\) is the magnetic flux through the loop. For a loop of radius 1 cm, the area (\(A\)) is: \(A = \pi \times (0.01)^2\).
4. Compute the flux and its change: The magnetic flux is \(\Phi = B \times A = \mu_0 \times 500 \times 10\sin(1000t) \times \pi \times (0.01)^2\). So, the time derivative, required for the emf calculation, involves the derivative of \(\sin(1000t)\): \(\varepsilon = - \frac{d}{dt} [\mu_0 \times 500 \times 10\sin(1000t) \times \pi \times (0.01)^2]\).
5. Calculate the induced emf: \(\varepsilon = - \mu_0 \times 500 \times 10 \times \pi \times (0.01)^2 \times \frac{d}{dt} [\sin(1000t)]\).
   This gives: \(\varepsilon = - \mu_0 \times 500 \times 10 \times \pi \times (0.01)^2 \times 1000 \times \cos(1000t)\).
6. Determine the r.m.s. value of the current: The r.m.s. value of the current \(I_{\text{rms}}\) through the 10 \(\Omega\) loop is \(I_{\text{rms}} = \frac{\varepsilon_{\text{rms}}}{R}\). 
   Average out the squared cos component, assuming over a cycle to simplify: \(\varepsilon_{\text{rms}} = \mu_0 \times 500 \times 10 \times \pi \times (0.01)^2 \times 1000 / \sqrt{2}\).
7. Conclusion and Final Calculation: From the relationship given: \(I_{\text{rms}} = \frac{\alpha}{\sqrt{2} \mu A}\), equate the expressions for \(I_{\text{rms}}\) to solve for \(\alpha\). Solve this equation numerically to give us \(I_{rms} \approx 197\), leading to the determination of \(\alpha = 197\).

Therefore, the value of \(\alpha\) is 197.