If number of turns and current becomes double for any solenoid, then value of magnetic field becomes

Options

A. twice
B. same
C. half
D. four times

Answer

To solve the problem, we need to analyze how the magnetic field inside a solenoid changes when both the number of turns and the current are doubled.

### Step-by-Step Solution:

1. **Understand the Formula for Magnetic Field in a Solenoid**:
The magnetic field \( B \) inside a solenoid is given by the formula:
\[ B = \mu_0 \cdot n \cdot I \]
where:
- \( \mu_0 \) is the magnetic permeability of free space,
- \( n \) is the number of turns per unit length,
- \( I \) is the current flowing through the solenoid.

2. **Define the Number of Turns per Unit Length**:
If the total number of turns in the solenoid is \( N \) and the length of the solenoid is \( L \), then the number of turns per unit length \( n \) can be expressed as:
\[ n = \frac{N}{L} \]

3. **Substitute \( n \) into the Magnetic Field Formula**:
Substituting \( n \) into the magnetic field formula gives:
\[ B = \mu_0 \cdot \left(\frac{N}{L}\right) \cdot I \]

4. **Consider the Changes in Number of Turns and Current**:
If the number of turns \( N \) is doubled, the new number of turns \( N' \) becomes:
\[ N' = 2N \]
If the current \( I \) is also doubled, the new current \( I' \) becomes:
\[ I' = 2I \]

5. **Calculate the New Magnetic Field**:
The new magnetic field \( B' \) can be calculated using the new values:
\[ B' = \mu_0 \cdot \left(\frac{N'}{L}\right) \cdot I' = \mu_0 \cdot \left(\frac{2N}{L}\right) \cdot (2I) \]
Simplifying this, we get:
\[ B' = \mu_0 \cdot \frac{2N}{L} \cdot 2I = 4 \cdot \left(\mu_0 \cdot \frac{N}{L} \cdot I\right) = 4B \]

6. **Conclusion**:
Therefore, the new magnetic field \( B' \) is four times the original magnetic field \( B \):
\[ B' = 4B \]

### Final Answer:
The value of the magnetic field becomes **4 times the original magnetic field**.