Question

What is the solenoid ? Obtain the expression for magnetic field inside a long current carrying solenoid.

Hint:

The formula \(B = \mu_0 n I\) is for the field deep inside a long solenoid. Near the ends of a finite solenoid, the field is weaker and non-uniform. The field at the very end of a long solenoid is approximately half the value of the field at its center, i.e., \(B_{end} \approx \frac{1}{2}\mu_0 n I\).

Answer 1

Step 1: Definition of a Solenoid: 
A solenoid is a long coil of wire consisting of many closely wound turns, usually in the shape of a helix. When an electric current flows through the wire, the solenoid acts as an electromagnet. A "long" or "ideal" solenoid is one whose length is much greater than its diameter. Inside a long solenoid, the magnetic field is strong, uniform, and directed along the axis of the solenoid. The magnetic field outside is negligibly weak. 
Step 2: Derivation using Ampere's Circuital Law: 
We can derive the expression for the magnetic field (\(\vec{B}\)) inside a long solenoid using Ampere's Circuital Law, which states: \[ \oint \vec{B} . d\vec{l} = \mu_0 I_{\text{enclosed}} \] Consider a long solenoid with \(n\) turns per unit length carrying a current \(I\). We choose a rectangular Amperian loop PQRS of length L as shown in the diagram, with side PQ inside the solenoid and parallel to its axis. 

Step 3: Evaluating the Line Integral: 
The line integral of the magnetic field around the closed loop PQRS is the sum of the integrals over its four segments: \[ \oint_{PQRS} \vec{B} . d\vec{l} = \int_{P}^{Q} \vec{B} . d\vec{l} + \int_{Q}^{R} \vec{B} . d\vec{l} + \int_{R}^{S} \vec{B} . d\vec{l} + \int_{S}^{P} \vec{B} . d\vec{l} \] 
Along PQ: The magnetic field \(\vec{B}\) is uniform and parallel to the path element \(d\vec{l}\). So, \( \int_{P}^{Q} \vec{B} . d\vec{l} = \int_{P}^{Q} B \, dl \cos(0^\circ) = B \int_{P}^{Q} dl = BL \). 
Along QR and SP: The magnetic field \(\vec{B}\) is perpendicular to the path elements \(d\vec{l}\). So, \( \vec{B} . d\vec{l} = B \, dl \cos(90^\circ) = 0 \). Thus, the integrals are zero. 
Along RS: This segment is outside the long solenoid where the magnetic field is considered to be zero (\(B \approx 0\)). So, the integral is zero. Therefore, the total line integral is: \[ \oint \vec{B} . d\vec{l} = BL + 0 + 0 + 0 = BL \] Step 4: Finding the Final Expression: 
The total current enclosed by the loop, \(I_{\text{enclosed}}\), is the number of turns within the loop multiplied by the current in each turn. Number of turns in length L = \( nL \). \[ I_{\text{enclosed}} = (nL)I \] Applying Ampere's Law: \[ BL = \mu_0 (nLI) \] Canceling L from both sides, we get the expression for the magnetic field inside a long solenoid: \[ B = \mu_0 n I \]