Page:Elementary Text-book of Physics (Anthony, 1897).djvu/288

We assume that whatever bodies are in the field are of such a character that their magnetization is proportional to the magnetizing force; on this assumption, the potential at any point and the magnitude of the poles vary in the same proportion. Let $m_{1},m_{2},\dots m_{n}$ represent the values of the respective poles, and $V_{1},V_{2},\dots V_{n}$ the potentials at the places occupied by them in the final condition of the field. Each of these poles may be conceived of as an assemblage of a great number $n$ of small poles, each equal to ${\frac {m}{n}}\cdot$ If we think of the region occupied by the field as originally free from magnets, its energy after the magnets are present in it will be equal to the worls done in forming the magnet poles by the successive addition of such elementary poles. Let the field be free from magnetism, and let the quantities of magnetism ${\frac {m_{1}}{n}},{\frac {m_{2}}{n}},\dots {\frac {m_{n}}{n}},$ be brought to the points which the separate poles occupy in the final condition of the field; since the potentials at those points are originally zero, no work will be done in this operation. The presence of these poles causes a rise of potential throughout the field, and the potentials at the places occupied by the poles become ${\frac {V_{1}}{,}}{n}{\frac {V_{2}}{n}},\dots {\frac {V_{n}}{n}}\cdot$ Let elementary poles similar to those already introduced be brought to their respective places in the field; the work done on any one of them is ${\frac {mV}{n^{2}}},$ and the work done on them all is $\sum {\frac {mV}{n^{2}}}\cdot$ By this increase in the quantities at the poles the potentials become $2{\frac {V_{1}}{n}},2{\frac {V_{2}}{n}},\dots 2{\frac {V_{n}}{n}}\cdot$ This operation is repeated until $n$ quantities have been brought to each pole, so that the poles are in their final condition and the potential has everywhere its final value. The work done in bringing up the $n^{\text{th}}$ elementary pole to its place is ${\frac {m}{n}}(n-1){\frac {V}{n}};$ the work done in forming the field is therefore $\sum \left({\frac {1+2+3+\dots +(n-1)}{n^{2}}}\right)\cdot$ Now