A long solenoid carrying a current produces a magnetic field `B` along its axis. If the current is doubled and the number of turns per cm is halved, the new vlaue of the magnetic field is

Options

A. Both will follow a circular path with same radius
B. 2B
C. 4B
D. `B//2`

Answer

To solve the problem, we need to understand how the magnetic field \( B \) produced by a long solenoid is calculated. The formula for the magnetic field inside a long solenoid is given by:

\[ B = \mu_0 N I \]

where:
- \( B \) is the magnetic field,
- \( \mu_0 \) is the permeability of free space,
- \( N \) is the number of turns per unit length (in turns per meter),
- \( I \) is the current flowing through the solenoid.

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- Let the initial number of turns per unit length be \( N \) (in turns/meter).
- Let the initial current be \( I \).

Therefore, the initial magnetic field \( B \) can be expressed as:
\[ B = \mu_0 N I \]

2. **Determine the New Conditions**:
- The current is doubled, so the new current \( I' = 2I \).
- The number of turns per unit length is halved, so the new number of turns per unit length \( N' = \frac{N}{2} \).

3. **Calculate the New Magnetic Field**:
Using the formula for the magnetic field with the new values:
\[ B' = \mu_0 N' I' \]
Substituting the new values:
\[ B' = \mu_0 \left(\frac{N}{2}\right) (2I) \]

4. **Simplify the Expression**:
\[ B' = \mu_0 \left(\frac{N}{2} \cdot 2I\right) = \mu_0 N I \]
Therefore, we find that:
\[ B' = B \]

5. **Conclusion**:
The new magnetic field \( B' \) is equal to the original magnetic field \( B \). Thus, there is no change in the magnetic field strength.

### Final Answer:
The new value of the magnetic field is \( B' = B \).