Question

A small ring is given some velocity along the axis of a solenoid and it remains coaxial with solenoid. Current in solenoid is \(i = 10 \sin(\omega t)\); \(\omega = 1000\) rad/s. Number of turns per unit length is 500/m. Radius of ring is 1 cm and its resistance is 10\(\Omega\). Find RMS value of induced current in the ring :

• A. \(\sqrt{2} \times 10^{-5}\) A
• B. \(3 \times 10^{-4}\) A
• C. \(\sqrt{2} \times 10^{-4}\) A
• D. \(5 \times 10^{-6}\) A

Hint:

The velocity of the ring does not affect the induced EMF in this case because the magnetic field inside an ideal solenoid is uniform. The EMF is purely due to the time-varying current (transformer EMF). If the field were non-uniform, there would also be a motional EMF component.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We have a ring placed coaxially inside a solenoid. The solenoid has a time-varying current, which creates a time-varying magnetic field. This changing magnetic field, in turn, creates a changing magnetic flux through the ring, inducing an electromotive force (EMF) and a current in the ring according to Faraday's law of induction. We need to find the RMS value of this induced current.
Step 2: Key Formula or Approach:
1. Magnetic Field in a Solenoid: \(B = \mu_0 n i\), where \(n\) is the number of turns per unit length.
2. Magnetic Flux: \(\Phi = B \cdot A\), where A is the area of the ring.
3. Faraday's Law of Induction: \(\mathcal{E} = -\frac{d\Phi}{dt}\).
4. Ohm's Law: Induced current \(I_{ind} = \frac{\mathcal{E}}{R}\).
5. RMS Value: For a sinusoidal current \(I = I_0 \cos(\omega t)\), the RMS value is \(I_{rms} = \frac{I_0}{\sqrt{2}}\).
Step 3: Detailed Explanation:
Part A: Magnetic Flux through the Ring
The magnetic field inside the solenoid is \(B = \mu_0 n i = \mu_0 n (10 \sin(\omega t))\). This field is uniform inside the solenoid and parallel to the axis.
The area of the ring is \(A = \pi r^2\).
The magnetic flux through the ring is \(\Phi = B \cdot A = \mu_0 n (10 \sin(\omega t)) (\pi r^2)\).
Part B: Induced EMF
\[ \mathcal{E} = -\frac{d\Phi}{dt} = - \frac{d}{dt} \left[ 10 \pi r^2 \mu_0 n \sin(\omega t) \right] \] \[ \mathcal{E} = - 10 \pi r^2 \mu_0 n \frac{d}{dt}(\sin(\omega t)) = - 10 \pi r^2 \mu_0 n \omega \cos(\omega t) \] The peak value (amplitude) of the induced EMF is \(\mathcal{E}_0 = 10 \pi r^2 \mu_0 n \omega\).
Part C: Induced Current and its RMS Value
The induced current is \(I_{ind} = \frac{\mathcal{E}}{R} = -\frac{10 \pi r^2 \mu_0 n \omega}{R} \cos(\omega t)\).
The peak value of the induced current is \(I_0 = \frac{10 \pi r^2 \mu_0 n \omega}{R}\).
The RMS value of the induced current is \(I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{10 \pi r^2 \mu_0 n \omega}{\sqrt{2} R}\).
Part D: Calculation
Substitute the given values (in SI units):
- \(r = 1 \text{ cm} = 0.01 \text{ m}\)
- \(\mu_0 = 4\pi \times 10^{-7}\) T·m/A
- \(n = 500\) m\(^{-1}\)
- \(\omega = 1000\) rad/s
- \(R = 10\) \(\Omega\)
\[ I_{rms} = \frac{10 \pi (0.01)^2 (4\pi \times 10^{-7}) (500) (1000)}{\sqrt{2} \times 10} \] \[ I_{rms} = \frac{\pi (10^{-4}) (4\pi \times 10^{-7}) (500) (1000)}{\sqrt{2}} \] \[ I_{rms} = \frac{4\pi^2 \times (10^{-4} \times 10^{-7} \times 5 \times 10^2 \times 10^3)}{\sqrt{2}} = \frac{20\pi^2 \times 10^{-6}}{\sqrt{2}} \] Using the approximation \(\pi^2 \approx 10\):
\[ I_{rms} \approx \frac{20 \times 10 \times 10^{-6}}{\sqrt{2}} = \frac{200 \times 10^{-6}}{\sqrt{2}} = \frac{2 \times 10^{-4}}{\sqrt{2}} = \sqrt{2} \times 10^{-4} \text{ A} \] Step 4: Final Answer:
The RMS value of the induced current in the ring is \(\sqrt{2} \times 10^{-4}\) A.