中国科学技术大学：《电磁学》课程教学资源（PPT课件讲稿）Chapter 13 Magnetic Fields：VI

Chapter 13 Magnetic Fields: VI a Forces on a Wire Carrying a current in a Magnetic Field Magnetic Pressure Magnetic Energy denisity Magnetic Forces btwn TWo Electric Currents a Magnetic Forces within an Isolated Circuit

Chapter 13Magnetic Fields:VI ◼ Forces on a Wire Carrying a Current in a Magnetic Field ◼ Magnetic Pressure ◼ Magnetic Energy Denisity ◼ Magnetic Forces btwn Two Electric Currents ◼ Magnetic Forces Within an Isolated Circuit

In this Chapter, we study magnetic energy and macroscopic forces In practice, magnetic forces are much larger than electric forces and they have many more applications

In this Chapter, we study magnetic energy and macroscopic forces. In practice, magnetic forces are much larger than electric forces, and they have many more applications

13. 1 Force on a Wire Carrying a Current in a Mag- netic Field We know the Lorentz force: F=QVX B Now look at a piece dl of wire carrying current 1, a is its cross-section area, n is its density of electron Figure 13-1 Wire of cross-section a carrying a current /. The charges -e move at a velocity v. The magnetic field B is due to currents flowing elsewhere

Since the conduction electrons are moving the force on the wire is dF=md(c-ev×B), where nadl is the number of electrons in this piece of wire The current i in the wire is nev So dF=Ia×B=Idl×B Where dl is in the direction of the current If it's straight length l in a uniform of B, then F=n×B

Figure 13-2 Magnetic field near a current-carrying wire situated in a uniform magnetic field. (a) Lines of B for a current perpendicular to the paper. ( b)Lines of B for a uniform field parallel to the paper. (c) Superposition of fields in (a)and (b) and the resulting magnetic force F. The wire carres a current of 10 amperes, the uniform field has a B of 2 x 10"tesla, and the point where the lines of B are broken is at 10 millimeters from the center of the wire

In the above example the force on the current l is just given by F=lB and its direction is to the left The lines of b are plotted in fig 13-2 These lines are under tension and repel each other laterally, giving rise to magnetic forces

13.2 Magnetic Pressure a current sheet carries a current density a(am pere/meter). It generates B Bv ∮B·dl=poI, taking a closed path around the current, we have B+B=0. so the magnetic field generated on both Sides of the sheet is B 0 its direction one side is upward another side is downward(both⊥toa)

13.2 Magnetic Pressure Now we put this sheet into a given uniform magnetic field=p0I 2 Thus on one side of the sheet the magnetic fields cance (0 and on the other side they add up 10/2+p0/2=p0l

13.2 Magnetic Pressure Now we put this sheet into a given uniform magnetic field B′=p0O Thus on one side of the sheet the magnetic fields cancel, 10c/2-10/2=0 and on the other side they add up 40/2+140/2=10=2B The force on the sheet per unit area( pressure)is =axB′=a(0/2)=a210/2=(2B220

Conclusion: for a sheet carrying a current with a total magnetic field b on one side and zero mag- netic field on the other side the sheet is subject to a pressure B 0 Its direction is the same as if the b is replaced by a compressed gas. The magnetic pressure here can be ascribed to the lateral repulsion of the magnetic lines