A long solenoid has n turns per metre and current IA is flowing through it. The magnetic field induciton at the ends of the solenoid is

Options

A. zero
B. `mu_0nI//2`
C. `mu_0nI`
D. `2mu_0NI`

Answer

To find the magnetic field induction at the ends of a long solenoid with \( n \) turns per meter and a current \( I_A \) flowing through it, we can follow these steps:

### Step-by-Step Solution:

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

2. **Determine the Magnetic Field at the Ends**:
At the center of the solenoid, the magnetic field is uniform and given by the formula above. However, at the ends of the solenoid, the magnetic field is not uniform and is weaker. It can be derived that the magnetic field at the ends of a long solenoid is half of that at the center:
\[ B_{\text{end}} = \frac{1}{2} B_{\text{center}} \]

3. **Substitute the Expression for \( B_{\text{center}} \)**:
We already know that:
\[ B_{\text{center}} = \mu_0 n I \]
Therefore, substituting this into our expression for \( B_{\text{end}} \):
\[ B_{\text{end}} = \frac{1}{2} (\mu_0 n I) \]

4. **Final Expression for the Magnetic Field at the Ends**:
Thus, the magnetic field induction at the ends of the solenoid can be expressed as:
\[ B_{\text{end}} = \frac{\mu_0 n I}{2} \]

### Final Answer:
The magnetic field induction at the ends of the solenoid is:
\[ B_{\text{end}} = \frac{\mu_0 n I_A}{2} \]