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. 2B
B. 4B
C. `B/2`
D. B

Answer

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

\[ B = \mu_0 N I \]

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), and
- \( I \) is the current flowing through the solenoid.

### Step 1: Initial Conditions
Let’s denote the initial number of turns per centimeter as \( N \) and the initial current as \( I \). Thus, the initial magnetic field \( B \) can be expressed as:

\[ B = \mu_0 N I \]

### Step 2: Changes in Current and Turns
According to the problem:
- The current is doubled: \( I' = 2I \)
- The number of turns per centimeter is halved: \( N' = \frac{N}{2} \)

### Step 3: Calculate New Magnetic Field
Now we can calculate the new magnetic field \( B' \) using the new values of current and turns:

\[ B' = \mu_0 N' I' = \mu_0 \left(\frac{N}{2}\right)(2I) \]

### Step 4: Simplifying the Expression
Substituting the values into the equation:

\[ B' = \mu_0 \left(\frac{N}{2}\right)(2I) = \mu_0 N I \]

### Step 5: Conclusion
Notice that \( \mu_0 N I \) is the original magnetic field \( B \). Therefore, we find that:

\[ B' = B \]

Thus, the new value of the magnetic field remains the same as the initial value.

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

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