2 Engineering Example 3

2.1 Magnetic flux

Introduction

The magnitude of the magnetic flux density on the axis of a solenoid, as in Figure 13, can be found by the integral:

$B={\int \; <!--nolimits\; -->}_{{\beta }_1}^{{\beta }_2}\frac{{\mu }_{0}nI}{2}sin\beta d\beta$

where ${\mu }_0$ is the permeability of free space ( $\approx 4\pi \times 10^{-7}\text{H}{\text{m}}^{-1}$ ), $n$ is the number of turns and $I$ is the current.

Figure 13:  

Problem in words

Predict the magnetic flux in the middle of a long solenoid.

Mathematical statement of the problem

We assume that the solenoid is so long that ${\beta }_1\approx 0$ and ${\beta }_2\approx \pi$ so that

$B={\int \; <!--nolimits\; -->}_{{\beta }_1}^{{\beta }_2}\frac{{\mu }_{0}nI}{2}sin\beta d\beta \approx {\int \; <!--nolimits\; -->}_0^{\pi }\frac{{\mu }_{0}nI}{2}sin\beta d\beta$

Mathematical analysis

The factor $\frac{{\mu }_{0}nI}{2}$ can be taken outside the integral i.e.

$B=\frac{{\mu }_{0}nI}{2}{\int \; <!--nolimits\; -->}_0^{\pi }sin\beta d\beta =\frac{{\mu }_{0}nI}{2}{\left[-cos\beta \right]}_0^{\pi }=\frac{{\mu }_{0}nI}{2}\left(-cos\pi +cos0\right)$

$=\frac{{\mu }_{0}nI}{2}\left(-\left(-1\right)+1\right)={\mu }_{0}nI$

Interpretation

The magnitude of the magnetic flux density at the midpoint of the axis of a long solenoid is predicted to be approximately ${\mu }_{0}nI$ i.e. proportional to the number of turns and proportional to the current flowing in the solenoid.