. A text book of physics, for the use of students of science and engineering . FIG. 899.—Magnetic field due toa straight current. Fig. 900.—Endless solenoid. Magnetic field inside an endless solenoid.—A circular ring uponwhich a wire is uniformly wound, so that the magnetic field iseverywhere in the direction of the circumference of the ring, is calledan endless solenoid. If there are n turns per centimetre length of thesolenoid, measured circumferentially, the total number of turns is2irrn, where r is the radius (Fig. 900). Thus, for a cv rent i in thewire, a unit magnetic pole carried round

. A text book of physics, for the use of students of science and engineering . FIG. 899.—Magnetic field due toa straight current. Fig. 900.—Endless solenoid. Magnetic field inside an endless solenoid.—A circular ring uponwhich a wire is uniformly wound, so that the magnetic field iseverywhere in the direction of the circumference of the ring, is calledan endless solenoid. If there are n turns per centimetre length of thesolenoid, measured circumferentially, the total number of turns is2irrn, where r is the radius (Fig. 900). Thus, for a cv rent i in thewire, a unit magnetic pole carried round Stock Photo
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. A text book of physics, for the use of students of science and engineering . FIG. 899.—Magnetic field due toa straight current. Fig. 900.—Endless solenoid. Magnetic field inside an endless solenoid.—A circular ring uponwhich a wire is uniformly wound, so that the magnetic field iseverywhere in the direction of the circumference of the ring, is calledan endless solenoid. If there are n turns per centimetre length of thesolenoid, measured circumferentially, the total number of turns is2irrn, where r is the radius (Fig. 900). Thus, for a cv rent i in thewire, a unit magnetic pole carried round the curved axis of thesolenoid passes 2irrn times round the current, and the work done istherefore ±ir{2irrni) ergs. But if H is the strength of magnetic field, 966 MAGNETISM AND ELECTRICITY chap. this is the force on the unit pole, and the work done upon the polein making a complete circuit along this path is 2tt;H ; .*. 2irrH=±ir(2irrni), or, H=47rm gauss (2) If the current be given in amperes, the strength of field is then 47ml /Q H =-|q- gauss 6> Unless the