A certain elastic conducting material is stretched into a circular loop. It is placed with its plane perpendicular to a uniform magnetic field B = 0.8 T. When released the radius of the loop starts shrinking at a constant rate of 2 cm/s. The induced emf in the loop at an instant when the radius of the loop is 10 cm will be _____ mV.
A certain elastic conducting material is stretched into a circular loop. It is placed with its plane perpendicular to a uniform magnetic field B = 0.8 T. When released the radius of the loop starts shrinking at a constant rate of 2 cm/s. The induced emf in the loop at an instant when the radius of the loop is 10 cm will be _____ mV.
Show Hint
For induced emf in circular loops:
- Use \(\mathcal{E} = -B \cdot 2\pi r \cdot \frac{dr}{dt}\) for shrinking loops.
- Ensure all units (e.g., \(r\), \(\frac{dr}{dt}\)) are consistent before calculations.
Correct Answer: 10
Solution and Explanation
1. Magnetic Flux: - The magnetic flux through the loop is:
\[ \Phi = B \cdot A = B \cdot \pi r^2, \]where \( B = 0.8 \, \text{T} \) and \( r = 10 \, \text{cm} = 0.1 \, \text{m} \).
2. Rate of Change of Flux: - The emf induced is:
\[ \mathcal{E} = -\frac{d\Phi}{dt}. \]- Differentiate \( \Phi \) with respect to time:
\[ \mathcal{E} = -\frac{d}{dt}(B \pi r^2) = -B \cdot 2 \pi r \frac{dr}{dt}. \]3. Substitute Values: - \( B = 0.8 \, \text{T}, r = 0.1 \, \text{m}, \frac{dr}{dt} = -2 \, \text{cm/s} = -0.02 \, \text{m/s}: \)
\[ \mathcal{E} = 0.8 \cdot 2 \pi \cdot 0.1 \cdot 0.02 = 0.010 \, \text{V}. \]4. Convert to mV:
\[ \mathcal{E} = 10 \, \text{mV}. \]Final Answer: 10 mV
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