Ampère’s Law (安培环路定理) In static magnetic field, if we integrate the magnetic field along an arbitrary loop, the resultant integral equals to the product of the permeability of the free space and the net current enclosed by the loop. (在真空中的稳恒磁场中,磁感应强度沿任意闭合回路的积分等 于真空磁导率乘以该回路所包围的电流的代数和) loop L B dl 0 I i i 1 Illustration of Ampère’s law See the right figure, a infinitely long current carrying wire with a current I. We choose a loop as shown in the figure and calculate the curve integral of the magnetic field: (如右图,一无穷长导线有电流 I 通过。选取如 图所示的环路并计算磁感应强度沿此曲线的积分可 得) 2 Accor di ng t o Bi ot - Savar t ' s l aw, t he magni t ude of t he magnet i c f i el d al ong t he l oop i s: 0 I B , and 2 r loop B dl loop Bdl loop Brd 0 I . 3 Generalization (普遍证明) As shown in the figure below, loop C1 circles an infinitely long current-carrying wire, and C2 circles no current, one integrate the magnetic field along the loops as following: 4 Circulation(C1 ) (环路C1的环量) B dl Bdl cos 2 0 C1 Brd C1 0 I 2 2 0 d 0 I Circulation(C2 ) Brd 0 (环路C2的环量) 5 (斯托克斯公式) B dl 0 I B 0 j j dS I loop Stokes formula B dl 0 I loop B 0 j 6 B dl 0 I loop B 0 j The curl of a magnetic field is current density vector. (磁场是有旋场,磁场的旋度是电流密度矢量) 7 Vector Potential (磁矢势) B 0 a vector A, satisfy B A A is called vector potential of the magnetic field. (从静磁场的散度为零可以知道,一定存在一个矢量场 使得磁场是该矢量场的旋度场,这个矢量场称为磁场的 矢势) 8 Applications of Ampère’s law Ampère’s law is valid only for steady current. In the electrostatic field, we apply Gauss’s law to evaluate the electric field due to symmetric charge distributions; we will now apply the Ampère’s law to evaluate the magnetic field of the systems of symmetry. It is important to choose a proper Gaussian surface in using Gauss’s law to evaluate the electric field. Similarly, it is important to choose a proper loop in applying the Ampère’s law to evaluate the magnetic field. 9 The loop is to satisfy one or more following conditions 1. The magnitude of the magnetic field can be argued by symmetry to be a constant over the path. 2. The dot product in path integral can be expressed as a simple algebra product because B and dl are parallel. 3. The dot product is zero since B are perpendicular to dl . 10 The magnetic field created by a long straight current-carrying wire (Example 26-4) A long, straight wire of radius R carries a current I 0 that is uniformly distributed through the cross section of the wire. Calculate the magnetic field at a distance r from the centre of the wire in the regions: r R and r R. 11 Solution: For r R, let us choose path 1 in the figure--a circle of radius r centered at the wire. From symmetry, we see that B must be a constant in magnitude and parallel to dl at every point on this circle. Because the total current passing through the plane of the circle is I 0 Ampere's law applied to the circular path gives: B dl B dl 2 rB 0 I 0 0 I 0 B (for r R) 2 r 12 consider the case r R. we choose the circular path 2. Because the current is uniform over the cross-section of the wire, the fraction of the current enclosed by the circle of radius r must equal to the ratio of the area enclosed by path 2 and the cross-sectional area of the wire: I0r 2 I r2 I I0 R2 R2 Following the same procedure as for circular path 1, we apply Ampere's law to path 2: Now B dl B dl 2 rB 0 I 0 I 0 r B (for r R ). 2 R 2 0 I 0 (for r R ) 2 r B 0 I 0 r (for r R ) 2 R 2 13 Magnetic field of a long, straight current-carrying solenoid We’ve calculated the magnetic field of points on the axis of a solenoid starting from Biot-Savart law and discuss the two special cases: the infinite long and semi-infinite long solenoid, the results are presented in the following. B 0 nI 2 cos 2 cos 1 infinitelongsolenoid : 0 nI 0 nI semi-infinitelongsolenoid : 2 14 When a solenoid’s turns are closed spaced and its length is large compared with its radius, it approaches the case of an ideal solenoid, the field outside of the solenoid is zero, and the field inside is uniform. We will use ideal solenoid as a simplification model for a real solenoid. 15 We can use Ampere's law to obtain an expression of the magnetic field inside an ideal solenoid. Consider a rectangular path of length l and width w as shown in the figure. From Ampere's law, we obtain: B dl Bl nlI 0 B 0 nI 16 The Magnetic Field Created by a Toroid A device called a toroid is often used to create an almost uniform magnetic field in some enclosed area. The device consists of a conducting wire wrapped around a ring (a torus) made of a non-conducting material. For a toroid having N closely spaced turns of wire and air in the torus, calculate the magnetic field in the region occupied by the torus, a distance r from the centre. 17 Reasoning: To calculate the field inside the toroid, we calculate the line integral of B dl over a circular path of radius r. By symmetry, we see the magnetic field is constant in magnitude on this circle and tangent to it, so B dl Bdl. Furthmore, note that the closed path surrounds a circular area through which N loops of wire pass, eahc of which carries a current I . The total current is thus NI . 18 Solution : Ampere's law applied to the circular path gives B dl B dl B 0 NI 0 NI B 0 nI 2 r Therefore, magnetic field inside a toroid varies as 1/r and hence is non-uniform. If r is large compared with the cross sectional radius of the torus, however, the field is almost uniform inside the toroid. For an ideal toroid, in which the turns are closed spaced, the magnitude of the magnetic field external to the toroid is zero. In reality, the turns of a toroid form a helix rather than circular loops. As a result a small field always exists external to the coin. 19 Magnetic Field of An Infinite Current-carrying Plane (Eg. 26-5) The current density flowing through an infinite current-carrying plane is j. Find the magnetic field in space. 1 B 0 j 2 20 Fundamentals of Magnetostatics (静磁学的基本原理) (毕奥-萨法尔定律) (叠加原理) (磁场) B dl 0 I loop B 0 j (高斯定理) (安培定理) 21 The End 22