A solenoid of radius 2.50cm has 400turns and a length of 20.0cm. The current in the coil is changing with time such that an emf of 75X10^(-6) is produced.
The rate of change of current in the solenoid is

Options

A. 38.0mA/s
B. 19.0mA/s
C. 21.0mA/s
D. 50.0mA/s

Answer

To solve the problem step by step, we will calculate the inductance of the solenoid and then use it to find the rate of change of current.

### Step 1: Calculate the Inductance of the Solenoid
The formula for the inductance \( L \) of a solenoid is given by:

\[ L = \frac{\mu_0 n^2 A}{l} \]

where:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{H/m} \) (permeability of free space),
- \( n \) is the number of turns per unit length,
- \( A \) is the cross-sectional area,
- \( l \) is the length of the solenoid.

#### Given:
- Total number of turns \( N = 400 \)
- Length of the solenoid \( l = 20.0 \, \text{cm} = 0.2 \, \text{m} \)
- Radius of the solenoid \( r = 2.5 \, \text{cm} = 0.025 \, \text{m} \)

#### Calculate \( n \):
\[ n = \frac{N}{l} = \frac{400}{0.2} = 2000 \, \text{turns/m} \]

#### Calculate the cross-sectional area \( A \):
\[ A = \pi r^2 = \pi (0.025)^2 = \pi \times 0.000625 \approx 0.0019635 \, \text{m}^2 \]

#### Substitute values into the inductance formula:
\[ L = \frac{(4\pi \times 10^{-7}) \times (2000)^2 \times (0.0019635)}{0.2} \]

Calculating this gives:
\[ L \approx \frac{(4\pi \times 10^{-7}) \times 4000000 \times 0.0019635}{0.2} \]
\[ L \approx \frac{(4\pi \times 10^{-7}) \times 7927.5}{0.2} \approx 1.97 \times 10^{-3} \, \text{H} \]

### Step 2: Use the Induced EMF to Find the Rate of Change of Current
The induced EMF \( e \) in the solenoid is related to the rate of change of current \( \frac{di}{dt} \) by the formula:

\[ e = L \frac{di}{dt} \]

Rearranging this gives:

\[ \frac{di}{dt} = \frac{e}{L} \]

#### Given:
- \( e = 75 \times 10^{-6} \, \text{V} \)
- \( L \approx 1.97 \times 10^{-3} \, \text{H} \)

#### Substitute the values:
\[ \frac{di}{dt} = \frac{75 \times 10^{-6}}{1.97 \times 10^{-3}} \]

Calculating this gives:
\[ \frac{di}{dt} \approx 38.0 \times 10^{-3} \, \text{A/s} = 38.0 \, \text{mA/s} \]

### Final Answer:
The rate of change of current in the solenoid is approximately \( 38 \, \text{mA/s} \).

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