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 step by step, we will first calculate the average rate of change of current (di/dt) and then use that to find the self-inductance (L) of the coil.

### Step 1: Determine the change in current (di)
The current changes from +5.0 A to -5.0 A. Therefore, the change in current (di) can be calculated as:

\[ di = final\ current - initial\ current = (-5.0\ A) - (5.0\ A) = -10.0\ A \]

### Step 2: Calculate the average rate of change of current (di/dt)
The time interval (dt) during which this change occurs is given as 0.2 seconds. The average rate of change of current (di/dt) is calculated as:

\[ \frac{di}{dt} = \frac{di}{dt} = \frac{-10.0\ A}{0.2\ s} = -50.0\ A/s \]

### Step 3: Use the induced emf to find self-inductance (L)
The induced emf (E) is given as 0.20 V. According to the formula for induced emf in terms of self-inductance:

\[ E = L \cdot \frac{di}{dt} \]

Rearranging this formula to solve for L gives:

\[ L = \frac{E}{\frac{di}{dt}} = \frac{0.20\ V}{-50.0\ A/s} \]

Ignoring the negative sign (as we are interested in magnitude):

\[ L = \frac{0.20}{50.0} = 0.004\ H \]

### Step 4: Convert Henry to Millihenry
To convert Henry to Millihenry:

\[ L = 0.004\ H = 4\ mH \]

### Final Answers
- Average di/dt = -50 A/s
- Self-inductance L = 4 mH