A 2.00-H inductor carries a steady current of 0.500 A. When the switch in the circuit is opened, the current is effectively zero after 10.0 ms. What is the average induced emf in the inductor during this time interval?
A 2.00-H inductor carries a steady current of 0.500 A. When the switch in the circuit is opened, the current is effectively zero after 10.0 ms. What is the average induced emf in the inductor during this time interval?
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![### Problem Statement:
A 2.00-H inductor carries a steady current of 0.500 A. When the switch in the circuit is opened, the current is effectively zero after 10.0 ms. What is the average induced emf in the inductor during this time interval?
### Explanation:
This question involves an inductor in an electrical circuit. When a switch is opened, the current through the inductor drops to zero over a short time interval (10.0 ms). To find the average induced electromotive force (emf), we can use the formula for the induced emf in an inductor:
\[ \text{emf} = -L \frac{\Delta I}{\Delta t} \]
where:
- \( L \) is the inductance (2.00-H),
- \( \Delta I \) is the change in current (0.500 A to 0 A, hence 0.500 A),
- \( \Delta t \) is the change in time (10.0 ms or 0.010 s).
### Calculation:
Substituting the given values into the formula:
\[ \text{emf} = -(2.00 \, \text{H}) \times \frac{0.500 \, \text{A}}{0.010 \, \text{s}} \]
\[ \text{emf} = -100 \, \text{V} \]
The negative sign indicates the direction of the induced emf according to Lenz's Law, which means the emf will oppose the change in current. Therefore, the average induced emf in the inductor during this time interval is 100 V.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F18a542a1-0899-4ebe-acd6-685d74e53e0e%2F69786753-0979-4ff0-bed4-ef6b3bd9e983%2F8oc8dl4_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
A 2.00-H inductor carries a steady current of 0.500 A. When the switch in the circuit is opened, the current is effectively zero after 10.0 ms. What is the average induced emf in the inductor during this time interval?
### Explanation:
This question involves an inductor in an electrical circuit. When a switch is opened, the current through the inductor drops to zero over a short time interval (10.0 ms). To find the average induced electromotive force (emf), we can use the formula for the induced emf in an inductor:
\[ \text{emf} = -L \frac{\Delta I}{\Delta t} \]
where:
- \( L \) is the inductance (2.00-H),
- \( \Delta I \) is the change in current (0.500 A to 0 A, hence 0.500 A),
- \( \Delta t \) is the change in time (10.0 ms or 0.010 s).
### Calculation:
Substituting the given values into the formula:
\[ \text{emf} = -(2.00 \, \text{H}) \times \frac{0.500 \, \text{A}}{0.010 \, \text{s}} \]
\[ \text{emf} = -100 \, \text{V} \]
The negative sign indicates the direction of the induced emf according to Lenz's Law, which means the emf will oppose the change in current. Therefore, the average induced emf in the inductor during this time interval is 100 V.
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