com www.GOALias.blogspot.com

www.GOALias.blogspot.comPhysics5.20 (a) B h= ( IN r )oµ0/2 cos45 = 0.39G(b) East to west (i.e., the needle will reverse its original direction).5.21 Magnitude of the other field=−2o1.2 × 10 × sin15= ×sin 45−34.4 10 To5.22 R =meVeB2m e× kinetic energy=eB= 11.3 mUp or down deflection = R (1-cosθ) where sinθ = 0.3/11.3. We getdeflection ≃ 4 mm.5.23 Initially, total dipole moment= 0.15 × 1.5 × 10 -23 × 2.0 ×10 24= 4.5 J T –1Use Curie’s Law m ∝ B/T to get the final dipole moment= 4.5 × (0.98/0.84) × (4.2/2.8)= 7.9 J T –1µrµo5.24 Use the formula B = NIwhere µ2πRr(relative permeability) to getB = 4.48 T.5.25 Of the two, the relation µ l= – (e/2m)l is in accordance with classicalphysics. It follows easily from the definitions of µ land l:µl= IA = ( e / T ) πr22πrl = mvr = m T2where r is the radius of the circular orbit which the electron of massm and charge (–e) completes in time T. Clearly, µl/ l = e/2 m.Since charge of the electron is negative (= –e), it is easily seen thatµ and l are antiparallel, both normal to the plane of the orbit.µ /2 l.Note µ s/S in contrast to µ / l is e/m, i.e.,Therefore, =−( e m)ltwice the classically expected value. This latter result (verifiedexperimentally) is an outstanding consequence of modern quantumtheory and cannot be obtained classically.lCHAPTER 62986.1 (a) Along qrpq(b) Along prq, along yzx