Module 3 : MAGNETIC FIELD
Lecture 15 : Biot- Savarts' Law
  Example 9 : Field of a solenoid on its axis

Consider a solenoid of $N$turns. The solenoid can be considered as stacked up circular coils. The field on the axis of the solenoid can be found by superposition of fields due to all circular coils. Consider the field at P due to the circular turns between $z$and $z+dz$from the origin, which is taken at the centre of the solenoid. The point P is at $z=d$. If $L$is the length of the solenoid, the number of turns within $z$and $z+dz$is $Ndz/L= ndz$, where $n$is the number of turns per unit length.

The magnitude of the field at P due to these turns is given by

\begin{displaymath}dB = \frac{\mu_0NI dz}{2L}\frac{a^2}{\left[a^2+(z-d)^2\right]^{3/2}}\end{displaymath}

The field due to each turn is along $\hat k$; hence the fields due to all turns simply add up. The net field is

\begin{displaymath}\vec B = \frac{\mu_0NI a^2}{2L}\int_{-L/2}^{L/2}\frac{dz}{\left[a^2+(z-d)^2 \right]^{3/2}} \hat k\end{displaymath}

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