Question

Deduce the expression of intensity of magnetic field produced inside long current carrying solenoid with the help of Ampere's law.

Hint:

Remember that \(n\) in the formula \(B = \mu_0 n I\) is the number of turns per unit length (\(n = N/L\)), not the total number of turns. This is a common point of confusion. The formula shows that the field inside a long solenoid is independent of its radius and the position within the solenoid.

Answer 1

Step 1: Statement of Ampere's Circuital Law:
Ampere's law states that the line integral of the magnetic field \(\vec{B}\) around any closed loop (called an Amperian loop) is equal to \(\mu_0\) times the total current \(I_{enc}\) passing through the area enclosed by the loop. \[ \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc} \]

Step 2: Derivation for a Long Solenoid:
Consider a long solenoid with \(n\) turns per unit length, carrying a current \(I\).
\begin{enumerate} \item Symmetry and Magnetic Field: For an ideal long solenoid, the magnetic field inside is strong, uniform, and directed along the axis of the solenoid. The magnetic field outside is negligibly weak (approximately zero).
\item Amperian Loop: To apply Ampere's law, we choose a rectangular Amperian loop `pqrs` of length \(L\), as shown in the cross-sectional diagram. The side `pq` is inside the solenoid, parallel to the axis, while the side `rs` is outside.
\begin{center} \begin{tikzpicture} % Solenoid turns (cross-section) \foreach \x in {0,1,2,3,4,5,6} { \node at (\x, 0.5) {$\otimes$}; % Current into page \node at (\x, -0.5) {$\odot$}; % Current out of page } \draw[gray] (-0.5, 0.8) -- (6.5, 0.8); \draw[gray] (-0.5, -0.8) -- (6.5, -0.8); % Magnetic field lines \foreach \y in {-0.3,0,0.3} { \draw[->, blue, thick] (-0.5, \y) -- (6.5, \y); } \node at (3, 0) [above, blue] {$\vec{B}$}; \node at (3, 1.2) [above, blue] {$\vec{B} \approx 0$}; % Amperian loop \draw[red, thick, ->] (1, -0.2) -- (5, -0.2) node[midway, below] {$p \to q$}; \draw[red, thick, ->] (5, -0.2) -- (5, 1.2) node[midway, right] {$q \to r$}; \draw[red, thick, ->] (5, 1.2) -- (1, 1.2) node[midway, above] {$r \to s$}; \draw[red, thick, ->] (1, 1.2) -- (1, -0.2) node[midway, left] {$s \to p$}; \node at (0.8, -0.2) [below left] {$p$}; \node at (5.2, -0.2) [below right] {$q$}; \node at (5.2, 1.2) [above right] {$r$}; \node at (0.8, 1.2) [above left] {$s$}; \draw[<->] (1,-1) -- (5,-1) node[midway, below] {$L$}; \end{tikzpicture} \end{center} \item Evaluate the Line Integral: We evaluate \(\oint \vec{B} \cdot d\vec{l}\) for the loop `pqrs`. \[ \oint_{pqrs} \vec{B} \cdot d\vec{l} = \int_{p}^{q} \vec{B} \cdot d\vec{l} + \int_{q}^{r} \vec{B} \cdot d\vec{l} + \int_{r}^{s} \vec{B} \cdot d\vec{l} + \int_{s}^{p} \vec{B} \cdot d\vec{l} \] \begin{itemize} \item Along pq (inside): \(\vec{B}\) is parallel to \(d\vec{l}\). So, \(\int_{p}^{q} \vec{B} \cdot d\vec{l} = \int BL \cos(0^\circ) dl = B \int_{0}^{L} dl = BL\). \item Along qr and sp (perpendicular): \(\vec{B}\) is perpendicular to \(d\vec{l}\). So, \(\int \vec{B} \cdot d\vec{l} = \int B dl \cos(90^\circ) = 0\). \item Along rs (outside): The magnetic field outside is zero (\(B=0\)). So, \(\int_{r}^{s} \vec{B} \cdot d\vec{l} = 0\). \end{itemize} The total value of the line integral is: \(\oint \vec{B} \cdot d\vec{l} = BL + 0 + 0 + 0 = BL\). \item Calculate Enclosed Current (\(I_{enc}\)): The number of turns within the loop of length \(L\) is \(N = nL\). Since each turn carries current \(I\), the total current enclosed by the loop is: \[ I_{enc} = N \times I = nLI \] \item Apply Ampere's Law: \[ \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc} \implies BL = \mu_0 (nLI) \] \item Solve for B: Canceling \(L\) from both sides gives the magnetic field inside the solenoid: \[ B = \mu_0 n I \] \end{enumerate}