1、Example 14.1 Magnetic field of a straight current. As shown in Fig 14.10 there is flowing a steady current I in circuit. Find the magnetic field at P produced by the current in a straight of the circuit of length L. the distance from that segment to P is r.,Or,Example 14.2 Magnetic field of a circul
2、ar current. There is a current carrying circular coil radius R and current I . Find the distribution of magnetic field the axis of coil.,Define the magnetic moment of a current carrying coil to be,= 2,At the center of the coil , x=0,Eq.(14.29)gives the magnetic field produced by a coil of magnetic m
3、oment on it axis. The field at other point can also be calculated. The result is that the distribution of induction lines of the magnetic field produced by a magnetic moment shown in Fig.14.13 is similar to that of the electric field lines of the electric field produced by an electric dipole shown i
4、n Fig.b(except in the location of the electric dipole and the coil),Example 14.3 The magnetic field along the axis of a current-carrying solenoid. As shown n Fig.14.4 there is a straight solenoid of length L ,radius R, would evenly with n turns of coil per unit length. A current I is flowing in the
5、coil . Find the distribution of magnetic field on the axis of the solenoid,Or,For a very long straight solenoid for any point on the axis in the solenoid, 1 =, 2 =0,The distribution of B-lines of a current-carrying solenoid in Fig 14.16, which is similar to that in Fig.14.8(c) demonstrated with iron
6、 chips. The magnetic field outside the solenoid is very weak and inside is essentially uniform, the longer the solenoid ,the more remarkable this feature,Example 14.4 The distribution of the magnetic field of an infinitely long cylindrical surface current. Assume the radius of the cross-section of t
7、he cylindrical surface be R with total current I flowing on the surface uniformly along the length of the cylinder,Example 14.5 The distribution of the magnetic field of a current-carrying toroid . A toroid is a circular solenoid as shown in Fig .14.22(a). Assume that the radius of the circular ring
8、 R and there are N turns of coil wound closed and evenly on the ring(Fig.14.22(b)with current I flowing in the coil.,(in the toroid),For any point outside the toroid ,take the circle C or C” coaxial with the toroid as the Ampere loop . As now =0, we get,(outside the toroid),Example 14.6 A parallel p
9、late capacitor is composed of two circular conducting plates of radius R=0.2m . Find the magnetic field between the plates at points r1=0.1m and r2=0.3m from the axis of the plates when the capacitor is being charged with current =10.(Neglect edge effect),The electric field between the plates is,Tak
10、e the circle of radius r1 as the Ampere loop C1 , the circulation of B1 is,And, 2 ,SUMMARY,1. Current density: where is the average velocity of the current carriers , i.e., the drift velocity. Current: Continuity equation of current: 2. A classical microscopic view of current in metal : The directio
11、nal motion of free electrons in metal is a series of accelerated motion in succession and each accelerated motion starts from zero directional velocity,Conductivity: Ohms law: differential form : Common form: Electromotive force :the work done by non-electrostatic force when charge passing through3.
12、 Magnetic induction : Definition equation Where is the force on the test charge moving with velocity by the magnetic field.,4.The Biot-Savart law” the magnetic field of a current element Superposition principle Theorem of continuity of magnetic flux Magnetic field of a long straight current = 0 2 wi
13、th field lines all circles perpendicular to and centered on the current line. Magnetic field in along straight current carrying solenoid= 0 with field lines parallel to axis. 5. Ampere circuital theorem(for stead currents):,Magnetic field of a current carrying toroid = 0 () =0( 6. Magnetic field associated with changing electric field (Maxwells law)General Ampere circuital theorem,
