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![**Problem: Calculating Self-Induced EMF in a Solenoid**
A solenoid has an inductance of 6 mH. If the current through the solenoid is increased at the rate of 5 A/s, what will be the self-induced emf?
**Solution Approach:**
To find the self-induced electromotive force (emf), we can use the formula:
\[ \text{emf} = -L \frac{dI}{dt} \]
Where:
- \( L \) is the inductance (in henrys)
- \( \frac{dI}{dt} \) is the rate of change of current (in amperes per second)
Given:
- \( L = 6 \, \text{mH} = 0.006 \, \text{H} \)
- \( \frac{dI}{dt} = 5 \, \text{A/s} \)
Substituting the values:
\[ \text{emf} = -0.006 \times 5 = -0.03 \, \text{V} \]
Thus, the self-induced emf is \(-0.03 \, \text{V}\).
**Explanation:**
- The negative sign indicates that the emf opposes the change in current, in accordance with Lenz's Law.
- This problem simplifies understanding of how inductance and rate of change in current relate to induced emf in solenoids.](https://content.bartleby.com/qna-images/question/fc44715e-e997-4081-aac6-58f75fd336c7/d9e84e91-9f12-44f1-8eed-484f383ee085/oan76hq_processed.png)
Transcribed Image Text:**Problem: Calculating Self-Induced EMF in a Solenoid**
A solenoid has an inductance of 6 mH. If the current through the solenoid is increased at the rate of 5 A/s, what will be the self-induced emf?
**Solution Approach:**
To find the self-induced electromotive force (emf), we can use the formula:
\[ \text{emf} = -L \frac{dI}{dt} \]
Where:
- \( L \) is the inductance (in henrys)
- \( \frac{dI}{dt} \) is the rate of change of current (in amperes per second)
Given:
- \( L = 6 \, \text{mH} = 0.006 \, \text{H} \)
- \( \frac{dI}{dt} = 5 \, \text{A/s} \)
Substituting the values:
\[ \text{emf} = -0.006 \times 5 = -0.03 \, \text{V} \]
Thus, the self-induced emf is \(-0.03 \, \text{V}\).
**Explanation:**
- The negative sign indicates that the emf opposes the change in current, in accordance with Lenz's Law.
- This problem simplifies understanding of how inductance and rate of change in current relate to induced emf in solenoids.
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