The number of turns in a long solenoid is 500. The area of cross-section of solenoid is `2xx10^(-3)m^(2)`. If the value of magnetic induction , on passing a current of 2 amp , through it is `5xx10^(-3)` Tesla, the magnitude of magnetic flux connected with it in Webers will be __

Options

A. `5xx10^(-3)`
B. `10^(-2)`
C. `10^(-5)`
D. `2.5`

Answer

To solve the problem of finding the magnitude of magnetic flux connected with a solenoid, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Number of turns in the solenoid (N) = 500
- Area of cross-section of the solenoid (A) = \(2 \times 10^{-3} \, m^2\)
- Magnetic induction (B) = \(5 \times 10^{-3} \, T\)

2. **Understand the Formula for Magnetic Flux:**
The magnetic flux (\(\Phi\)) through a solenoid can be calculated using the formula:
\[ \Phi = N \times B \times A \]
where:
- \(N\) = number of turns
- \(B\) = magnetic induction
- \(A\) = area of cross-section

3. **Substitute the Values into the Formula:**
Now, substituting the values we have:
\[ \Phi = 500 \times (5 \times 10^{-3}) \times (2 \times 10^{-3}) \]

4. **Calculate the Magnetic Flux:**
First, calculate the product of \(B\) and \(A\):
\[ B \times A = (5 \times 10^{-3}) \times (2 \times 10^{-3}) = 10 \times 10^{-6} = 1 \times 10^{-5} \, T \cdot m^2 \]
Now, multiply this result by \(N\):
\[ \Phi = 500 \times (1 \times 10^{-5}) = 5 \times 10^{-3} \, Wb \]

5. **Final Result:**
Therefore, the magnitude of the magnetic flux connected with the solenoid is:
\[ \Phi = 5 \times 10^{-3} \, Wb \]