A 2.00-H indductor carries a steady current of 0.500A. When the switch in the circuit is opened,the current is effectively zero after 10.0ms. What is the average induced emf in the inductor during this time interval?

Options

A. 100V
B. 150V
C. 133V
D. 200V

Answer

To find the average induced emf in the inductor during the time interval when the switch is opened, we can follow these steps:

### Step 1: Identify the given values
- Inductance (L) = 2.00 H
- Initial current (I1) = 0.500 A
- Final current (I2) = 0 A (effectively zero)
- Time interval (T) = 10.0 ms = 10 × 10^-3 s

### Step 2: Calculate the change in current (di)
The change in current (di) can be calculated as:
\[ di = I2 - I1 = 0 - 0.500 = -0.500 \, \text{A} \]

### Step 3: Use the formula for average induced emf (E)
The average induced emf (E) in the inductor can be calculated using the formula:
\[ E = -L \frac{di}{dt} \]
Where:
- L = inductance
- di = change in current
- dt = time interval

### Step 4: Substitute the values into the formula
Substituting the known values into the formula:
\[ E = -2.00 \, \text{H} \times \frac{-0.500 \, \text{A}}{10 \times 10^{-3} \, \text{s}} \]

### Step 5: Simplify the expression
Calculating the right side:
\[ E = -2.00 \times \frac{-0.500}{0.010} \]
\[ E = -2.00 \times -50 \]
\[ E = 100 \, \text{V} \]

### Step 6: State the final answer
The average induced emf in the inductor during the time interval is:
\[ \text{Average induced emf} = 100 \, \text{V} \]