A 2m long solenoid having 1000 turns produces a flux density of `3.14 xx 10^(-3) T`. Current in the solenoid will be:

Options

A. `1A`
B. `2A`
C. `5A`
D. `10A`

Explanation

To find the current in the solenoid, we can use the formula for the magnetic flux density \( B \) in a solenoid, which is given by:

\[ B = \frac{\mu_0 n I}{L} \]

Where:
- \( B \) is the magnetic flux density (in Tesla),
- \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{T m/A} \)),
- \( n \) is the number of turns per unit length (in turns/m),
- \( I \) is the current (in Amperes),
- \( L \) is the length of the solenoid (in meters).

Given:
- Length of the solenoid \( L = 2 \, \text{m} \),
- Total number of turns \( N = 1000 \),
- Magnetic flux density \( B = 3.14 \times 10^{-3} \, \text{T} \).

### Step 1: Calculate the number of turns per unit length \( n \)

\[ n = \frac{N}{L} = \frac{1000}{2} = 500 \, \text{turns/m} \]

### Step 2: Rearrange the formula to solve for current \( I \)

From the formula for \( B \):

\[ B = \frac{\mu_0 n I}{L} \]

We can rearrange it to find \( I \):

\[ I = \frac{B L}{\mu_0 n} \]

### Step 3: Substitute the known values into the equation

Substituting the values we have:

\[ I = \frac{(3.14 \times 10^{-3} \, \text{T}) \times (2 \, \text{m})}{(4\pi \times 10^{-7} \, \text{T m/A}) \times (500 \, \text{turns/m})} \]

### Step 4: Simplify the equation

Calculating the denominator:

\[ \mu_0 n = 4\pi \times 10^{-7} \times 500 = 2\pi \times 10^{-4} \, \text{T m/A} \]

Now substituting this back into the equation for \( I \):

\[ I = \frac{(3.14 \times 10^{-3}) \times 2}{2\pi \times 10^{-4}} \]

### Step 5: Cancel out terms

Since \( \pi \approx 3.14 \):

\[ I = \frac{(3.14 \times 10^{-3}) \times 2}{2 \times 3.14 \times 10^{-4}} \]

The \( 3.14 \) cancels out:

\[ I = \frac{2 \times 10^{-3}}{2 \times 10^{-4}} = \frac{2}{2} \times 10^{1} = 1 \times 10^{1} = 10 \, \text{A} \]

### Final Answer

Thus, the current in the solenoid is:

\[ I = 10 \, \text{A} \]