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www.GOALias.blogspot.com162PhysicsFIGURE 4.23 In the Bohr modelof hydrogen-like atoms, thenegatively charged electron isrevolving with uniform speedaround a centrally placedpositively charged (+Z e)nucleus. The uniform circularmotion of the electronconstitutes a current. Thedirection of the magneticmoment is into the plane of thepaper and is indicatedseparately by ⊗.(ii) is subject to torque like a magnetic needle. This led Ampere to suggestthat all magnetism is due to circulating currents. This seems to be partlytrue and no magnetic monopoles have been seen so far. However,elementary particles such as an electron or a proton also carry an intrinsicmagnetic moment, not accounted by circulating currents.4.10.3 The magnetic dipole moment of a revolving electronIn Chapter 12 we shall read about the Bohr model of the hydrogen atom.You may perhaps have heard of this model which was proposed by theDanish physicist Niels Bohr in 1911 and was a stepping stoneto a new kind of mechanics, namely, quantum mechanics.In the Bohr model, the electron (a negatively charged particle)revolves around a positively charged nucleus much as aplanet revolves around the sun. The force in the former caseis electrostatic (Coulomb force) while it is gravitational forthe planet-Sun case. We show this Bohr picture of the electronin Fig. 4.23.The electron of charge (–e) (e = + 1.6 × 10 –19 C) performsuniform circular motion around a stationary heavy nucleusof charge +Ze. This constitutes a current I, where,eI = (4.32)Tand T is the time period of revolution. Let r be the orbitalradius of the electron, and v the orbital speed. Then,2ðrT = (4.33)vSubstituting in Eq. (4.32), we have I = ev/2πr.There will be a magnetic moment, usually denoted by µ l,associated with this circulating current. From Eq. (4.28) itsmagnitude is, µ l= Iπr 2 = evr/2.The direction of this magnetic moment is into the planeof the paper in Fig. 4.23. [This follows from the right-handrule discussed earlier and the fact that the negatively chargedelectron is moving anti-clockwise, leading to a clockwise current.]Multiplying and dividing the right-hand side of the above expression bythe electron mass m e, we have,eµl=2me( mvr)ee= l2m[4.34(a)]eHere, l is the magnitude of the angular momentum of the electronabout the central nucleus (“orbital” angular momentum). Vectorially,eµl=− l2m[4.34(b)]eThe negative sign indicates that the angular momentum of the electronis opposite in direction to the magnetic moment. Instead of electron with