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The magnetic field on the axis of a lon...

The magnetic field on the axis of a long solenoid having n turns per unit length and carrying a current is

A

`mu_(0)ni`

B

`mu_(0)n^(2)i`

C

`mu_(0)ni^(2)`

D

none of these

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The correct Answer is:
To find the magnetic field on the axis of a long solenoid with \( n \) turns per unit length and carrying a current \( I \), we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Solenoid**: A solenoid is a long coil of wire that generates a magnetic field when an electric current passes through it. The magnetic field inside a long solenoid is uniform and parallel to the axis of the solenoid. 2. **Define the Parameters**: - Let \( n \) be the number of turns per unit length of the solenoid. - Let \( I \) be the current flowing through the solenoid. 3. **Applying Ampere's Circuital Law**: According to Ampere's Circuital Law, the line integral of the magnetic field \( \mathbf{B} \) around a closed loop is equal to \( \mu_0 \) times the total current \( I_{\text{enc}} \) passing through the loop: \[ \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{enc}} \] 4. **Choosing an Amperian Loop**: We choose a rectangular Amperian loop that runs inside the solenoid along its axis and extends outside it. The magnetic field inside the solenoid is uniform, while outside it is negligible. 5. **Calculating the Current Enclosed**: The total current enclosed by the Amperian loop is: \[ I_{\text{enc}} = n \cdot I \] where \( n \) is the number of turns per unit length and \( I \) is the current through each turn. 6. **Evaluating the Integral**: Inside the solenoid, the magnetic field \( B \) is uniform, and the length of the path inside the solenoid is \( l \). Thus, we can write: \[ B \cdot l = \mu_0 (n \cdot I) \] where \( l \) is the length of the solenoid. 7. **Solving for the Magnetic Field**: Rearranging the equation gives us: \[ B = \frac{\mu_0 n I}{l} \] Since we are considering a long solenoid, we can simplify this to: \[ B = \mu_0 n I \] ### Final Result: The magnetic field on the axis of a long solenoid is given by: \[ B = \mu_0 n I \]

To find the magnetic field on the axis of a long solenoid with \( n \) turns per unit length and carrying a current \( I \), we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Solenoid**: A solenoid is a long coil of wire that generates a magnetic field when an electric current passes through it. The magnetic field inside a long solenoid is uniform and parallel to the axis of the solenoid. 2. **Define the Parameters**: ...
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