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www.GOALias.blogspot.comMagnetism andMatter2µ0ndxIadB =2[( r − x) + a ]2 232The magnitude of the total field is obtained by summing over all theelements — in other words by integrating from x = – l to x = + l. Thus,20nIaB µl dx= ∫ 2 2 3/22 − l[( r − x) + a ]This integration can be done by trigonometric substitutions. Thisexercise, however, is not necessary for our purpose. Note that the rangeof x is from – l to + l. Consider the far axial field of the solenoid, i.e.,r >> a and r >> l . Then the denominator is approximated by32 2 2 3[( r − x) + a ] ≈randB2 lµ 0 nIa=32r∫−ldx2µ0nI 2la=3(5.1)2 rNote that the magnitude of the magnetic moment of the solenoid is,m = n (2l) I (πa 2 ) — (total number of turns × current × cross-sectionalarea). Thus,µ0 2mB =3(5.2)4πrThis is also the far axial magnetic field of a bar magnet which one mayobtain experimentally. Thus, a bar magnet and a solenoid produce similarmagnetic fields. The magnetic moment of a bar magnet is thus equal tothe magnetic moment of an equivalent solenoid that produces the samemagnetic field.Some textbooks assign a magnetic charge (also called pole strength)+q mto the north pole and –q mto the south pole of a bar magnet of length2l, and magnetic moment q m(2l). The field strength due to q mat a distancer from it is given by µ 0q m/4πr 2 . The magnetic field due to the bar magnetis then obtained, both for the axial and the equatorial case, in a manneranalogous to that of an electric dipole (Chapter 1). The method is simpleand appealing. However, magnetic monopoles do not exist, and we haveavoided this approach for that reason.5.2.3 The dipole in a uniform magnetic fieldThe pattern of iron filings, i.e., the magnetic field lines gives us anapproximate idea of the magnetic field B. We may at times be required todetermine the magnitude of B accurately. This is done by placing a smallcompass needle of known magnetic moment m and moment of inertia Iand allowing it to oscillate in the magnetic field. This arrangement is shownin Fig. 5.4(b).The torque on the needle is [see Eq. (4.29)],τ = m × B (5.3)177