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www.GOALias.blogspot.comPhysics180EXAMPLE 5.4 EXAMPLE 5.3one of the two bars is a magnet, pick up one, (say, A) and lower one ofits ends; first on one of the ends of the other (say, B), and then on themiddle of B. If you notice that in the middle of B, A experiences noforce, then B is magnetised. If you do not notice any change from theend to the middle of B, then A is magnetised.5.2.4 The electrostatic analogComparison of Eqs. (5.2), (5.3) and (5.6) with the corresponding equationsfor electric dipole (Chapter 1), suggests that magnetic field at largedistances due to a bar magnet of magnetic moment m can be obtainedfrom the equation for electric field due to an electric dipole of dipole momentp, by making the following replacements:1 µ0E → B, p → m , →4πε04πIn particular, we can write down the equatorial field (B E) of a bar magnetat a distance r, for r >> l, where l is the size of the magnet:µ0mBE=−34 π r(5.7)Likewise, the axial field (B A) of a bar magnet for r >> l is:µ0 2mBA=3(5.8)4 π rEquation (5.8) is just Eq. (5.2) in the vector form. Table 5.1 summarisesthe analogy between electric and magnetic dipoles.TABLE 5.1 THE DIPOLE ANALOGYElectrostaticsMagnetism1/ε 0µ 0Dipole moment p mEquatorial Field for a short dipole –p/4πε 0r 3 – µ 0m / 4π r 3Axial Field for a short dipole 2p/4πε 0r 3 µ 02m / 4π r 3External Field: torque p × E m × BExternal Field: Energy –p.E –m.BExample 5.4 What is the magnitude of the equatorial and axial fieldsdue to a bar magnet of length 5.0 cm at a distance of 50 cm from itsmid-point? The magnetic moment of the bar magnet is 0.40 A m 2 , thesame as in Example 5.2.Solution From Eq. (5.7)BEµ m=410 × 0.4 10 × 0.4= = −70.125 = 3.2 × 10 T−7 −7033π r ( 0.5)From Eq. (5.8),BAµ02m=34πr= ×−76.4 10 T