中国科学技术大学：《电磁学》课程教学资源（PPT课件讲稿）Chapter 8 Magnetic Fields：I

Chapter 8 Magnetic Fields: I Magnetic Induction The divergence of B Magnetic Monopoles The vector potential a a and v Integral of a over a Closed curve

Chapter 8 Magnetic Fields:I ◼ Magnetic Induction ◼ The Divergence of B ◼ Magnetic Monopoles ◼ The Vector Potential A ◼ A and V ◼ Integral of A over a Closed Curve

8.1 The Magnetic Induction B ee FIg 8-1 a circuit carrying a current I generates the magnetic induction at the point Pas B X ri 4丌 where the integration is carried out over the circuit C and the constant po is called the permeability of free space. 10=4丌×10- tesla meter/ ampere Remark: the unit vector ri points from the source to the point of observation, i. e. from dl to P. b has unit tesla

dB dI Figure 8-l The magnetic induction dB=(Ho/4x)/dI x r/r produced by an element i dl of the current I in a circuit

If the current is distributed in space with a current den- sity J, then the element of current Idl is replaced by JdT, and the magnetic induction expressed as r B 4丌 Here the integration is carried over the volume occu pied by the currents Remarks: These two expressions for b are equivalent The flux of the magnetic induction b through a surface s is defined as =/sB·da Φ has unit weber

Example: Long Straight Wire Carrying a Current See Fig. 8-3 dR dI igure 8-3 The magnetic induction dB produced by an clement I dI of the current I in a long straight wire

Example: Long Straight Wire Carrying a Current See Fig. 8-3

An element dl with current I produces a magnetic in- duction dB=Addl×t=0 i sin edl 4丌 4丌 where 0 is the angle between dl and ri, and ol unit vector in the azimuthal direction Now using sin Adl cos adl=rda and p=r cos a, sin odl rda da cos ada 2 Thus integration yields the magnetic induction B= m/2 cos ada (0 4丌-7 P1 2TP See「ig8-4

Figure 8-4 Lines of B in a plane perpendicular to a long straight wire carrying a current i. The density of the lines is inversely proportional to the distance to the wire Lines close to the wire are not shown

Example 2 Circular Loop Magnetic Dipole Moment See Fig. 8-5 for a loop of radius a wit a current l We want to know the component of b along the z axis produced by the loop dB=0,d×r1.dBz4rr2, 0,dcos日 4丌 4O 2Ta cos 0 B7 T llo, a cos 0 By cos 0=a/r, and r2=z2+a2, we have Ia B 2(2+a2)1/2

dB Figure &-5 The magnetic induction dB produced by an clement I dl at a point on the axis of circular current loop of radius a. The projection of dB on the axis is dB

Figure 8-6 Lines of B in a plane containing the axis of the loop of Fig.8-5. The direction of the current in the loop is shown by the dot and the cross