Question:
A circular loop of area \(5~\text{mm}^2\) is placed coaxially inside a long solenoid that has 1000 turns/cm and carries a sinusoidally varying current of amplitude 1 A and angular frequency \(\omega\). If the amplitude of the emf induced in the loop is \(4.4 \times 10^{-4}~\text{V}\), then the value of \(\omega\) is:
A circular loop of area \(5~\text{mm}^2\) is placed coaxially inside a long solenoid that has 1000 turns/cm and carries a sinusoidally varying current of amplitude 1 A and angular frequency \(\omega\). If the amplitude of the emf induced in the loop is \(4.4 \times 10^{-4}~\text{V}\), then the value of \(\omega\) is:
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Use Faraday's law for induced emf in a small loop inside a solenoid: \(\mathcal{E} = N A \mu_0 n I_0 \omega\). Be careful with unit conversions.
Updated On: Jun 19, 2026
- 350 rad/s
- 700 rad/s
- 300 \(\pi\) rad/s
- 200 \(\pi\) rad/s
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The Correct Option is B
Solution and Explanation
Step 1: EMF induced in a coil.
The induced emf in a loop is given by Faraday’s law: \[ \mathcal{E} = N A \frac{dB}{dt} \] For sinusoidal current \(I = I_0 \sin \omega t\), \(\frac{dB}{dt} = \mu_0 n I_0 \omega \cos \omega t\). Amplitude: \(\mathcal{E}_0 = \mu_0 n I_0 \omega A\).
Step 2: Convert turns/cm to turns/m.
\[ n = 1000~\text{turns/cm} = 1000 \times 100 = 10^5~\text{turns/m} \]
Step 3: Substitute values.
\[ \mathcal{E}_0 = \mu_0 n I_0 A \omega \Rightarrow 4.4 \times 10^{-4} = (4 \pi \times 10^{-7}) (10^5)(1)(5 \times 10^{-6}) \omega \]
Step 4: Solve for \(\omega\).
\[ \omega = \frac{4.4 \times 10^{-4}}{(4 \pi \times 10^{-7}) (10^5)(5 \times 10^{-6})} \approx 700~\text{rad/s} \]
Step 5: Conclusion.
The angular frequency \(\omega = 700~\text{rad/s}\).
The induced emf in a loop is given by Faraday’s law: \[ \mathcal{E} = N A \frac{dB}{dt} \] For sinusoidal current \(I = I_0 \sin \omega t\), \(\frac{dB}{dt} = \mu_0 n I_0 \omega \cos \omega t\). Amplitude: \(\mathcal{E}_0 = \mu_0 n I_0 \omega A\).
Step 2: Convert turns/cm to turns/m.
\[ n = 1000~\text{turns/cm} = 1000 \times 100 = 10^5~\text{turns/m} \]
Step 3: Substitute values.
\[ \mathcal{E}_0 = \mu_0 n I_0 A \omega \Rightarrow 4.4 \times 10^{-4} = (4 \pi \times 10^{-7}) (10^5)(1)(5 \times 10^{-6}) \omega \]
Step 4: Solve for \(\omega\).
\[ \omega = \frac{4.4 \times 10^{-4}}{(4 \pi \times 10^{-7}) (10^5)(5 \times 10^{-6})} \approx 700~\text{rad/s} \]
Step 5: Conclusion.
The angular frequency \(\omega = 700~\text{rad/s}\).
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