An average induced emf of 0.20 V appears in a coil when the current in it is changed from 5.0A in one direction to 5.0 A in the opposite direction in 0.2 sec.

Options

A. Average `(di)/(dt)` will be equal to -50A/sec
B. Average `(di)/(dt)` will be equal to 50A/sec
C. The self inductance of the coil is 4.0 mH
D. The self inductance of the coil is 3.5 mH

Answer

To solve the problem, we need to determine the average rate of change of current (di/dt) and the self-inductance (L) of the coil based on the given information.

### Step-by-Step Solution

1. **Identify the change in current (ΔI)**:
The current changes from +5.0 A to -5.0 A. Therefore, the total change in current is:
\[ \Delta I = -5.0 \, \text{A} - 5.0 \, \text{A} = -10.0 \, \text{A} \]

2. **Calculate the time interval (Δt)**:
The time interval during which this change occurs is given as:
\[ \Delta t = 0.2 \, \text{s} \]

3. **Calculate the average rate of change of current (di/dt)**:
The average rate of change of current can be calculated using the formula:
\[ \frac{di}{dt} = \frac{\Delta I}{\Delta t} \]
Substituting the values we found:
\[ \frac{di}{dt} = \frac{-10.0 \, \text{A}}{0.2 \, \text{s}} = -50.0 \, \text{A/s} \]
(The negative sign indicates the direction of change, but we will consider the magnitude for further calculations.)

4. **Use the formula for induced emf (E)**:
The induced emf in the coil is related to the self-inductance (L) and the rate of change of current by the formula:
\[ E = -L \frac{di}{dt} \]
Given that the average induced emf (E) is 0.20 V, we can substitute the values:
\[ 0.20 \, \text{V} = -L \cdot (-50.0 \, \text{A/s}) \]
Simplifying this gives:
\[ 0.20 \, \text{V} = L \cdot 50.0 \, \text{A/s} \]

5. **Solve for self-inductance (L)**:
Rearranging the equation to find L:
\[ L = \frac{0.20 \, \text{V}}{50.0 \, \text{A/s}} = 0.004 \, \text{H} = 4 \, \text{mH} \]

### Final Answers
- The average rate of change of current (di/dt) is **50 A/s**.
- The self-inductance (L) of the coil is **4 mH**.