Ampere s Law. Outline. Objectives. BEE-Lecture Notes Anurag Srivastava 1

Transcription

1 Outline Introduce as an analogy to Gauss Law. Define. Applications of. Objectives Recognise to be analogous to Gauss Law. Recognise similar concepts: (1) draw an imaginary shape enclosing the current carrying conductor, (2) current enclosed. Anurag Srivastava 1

2 Learning Goals How you can determine the amount of charge within a closed surface by examining the electric field on the surface! What is meant by electric flux and how you can calculate it. How to use Gauss s Law to calculate the electric field due to a symmetric distribution of charges. A charge inside a box can be probed with a test charge q o to measure E field outside the box. The volume (V) flow rate (dv/dt) of fluid through the wire rectangle (a) is va when the area of the rectangle is perpendicular to the velocity vector v and (b) is va cos φ when the rectangle is tilted at an angle φ. We will next replace the fluid velocity flow vector v with the electric field vector E to get to the concept of electric flux Φ E. Volume flow rate through the wire rectangle. Anurag Srivastava 2

3 (a) The electric flux through the surface = EA. (b) When the area vector makes an angle φ with the vector E, the area projected onto a plane oriented perpendicular to the flow is A perp. = A cos φ. The flux is zero when φ = 90 o because the rectangle lies in a plane parallel to the flow and no fluid flows through the rectangle A flat surface in a uniform electric field. Φ E = E. da = E da cos φ = E da = E da = E (4π R 2 ) = (1/4π ε o ) q /R 2 ) (4π R 2 ) = q / ε o. So the electric flux Φ E = q / ε o. Now we can write Gauss's Law: Φ E = E. da = EdA cos φ =Qencl /ε o Electric FLUX through a sphere centered on a point charge q. The projection of an element of area da of a sphere of radius R onto a concentric sphere of radius 2R. The projection multiplies each linear dimension by 2, so the area element on the larger sphere is 4 da. The same number of lines of flux pass thru each area element. Flux Φ E from a point charge q. Anurag Srivastava 3

4 The projection of the area element da onto the spherical surface is da cos φ. Flux through an irregular surface. Spherical Gaussian surfaces around (a) positive and (b) negative point charge. Gauss s Law can be used to calculate the magnitude of the E field vector: Anurag Srivastava 4

5 Learning Goals How to calculate the magnetic field produced by a long straight current-carrying wire, using Law of Biot & Savart (the Biot Savart law is an equation describing the magnetic field generated by an electric current) How to calculate the magnetic field produced by a circular current-carrying loop of wire, using Law of Biot & Savart. How to use to calculate the magnetic field caused by symmetric current distributions. (a) Magnetic field caused by the current element Idl. (b) In figure (b) the current is moving into the screen. Magnetic field around a long, straight conductor. The field lines are circles, with directions determined by the right-hand rule. Anurag Srivastava 5

6 x o Magnetic field produced by a straight currentcarrying wire of length 2a. The direction of B at point P is into the screen. Law of Biot and Savart db = µ o / 4π (I dl x r) / r 3 Use Law of Biot and Savart, the integral is simple! db = µ o / 4π (I dl x r) / r 3 Magnetic field caused by a circular loop of current. The current in the segment dl causes the field db, which lies in the xy plane. Parallel conductors carrying currents in the same direction attract each other. The force on the upper conductor is exerted by the magnetic field caused by the current in the lower conductor. Anurag Srivastava 6

7 states that the integral of B around any closed path equals µ o times the current, I encircled, encircled by the closed loop. We will use this law to obtain some useful results by choosing a simple path along which the magnitude of B is constant, (or independent of dl). Some () integration paths for the line integral of B in the vicinity of a long straight conductor. Path in (c) is not useful because it does not encircle the current-carrying conductor. To find the magnetic field at radius r < R, we apply to the circle (path) enclosing the red area. For r > R, the circle (path) encloses the entire conductor. Anurag Srivastava 7

8 B = µ o n I, where n = N / L A section of a long, tightly wound solenoid centered on the x-axis, showing the magnetic field lines in the interior of the solenoid and the current. COAXIAL CABLE A solid conductor with radius a is insulated from a conducting rod with inner radius b and outer radius c. Gauss law allowed us to find the net electric field due to any charge distribution (with little effort) by applying symmetry. Similarly the net magnetic field can be found with little effort if there is symmetry using Ampere s law. Anurag Srivastava 8

9 Ampere s law, v v B. ds = µ 0i enc Where the integral is a line integral. B.ds is integrated around a closed loop called an Amperian loop. The current i enc is net current enclosed by the loop. ie, ie i enc N v v B. ds = µ 0 i n n= 1 = N i n n= 1 Example. Find the magnetic field outside a long straight wire with current. r I Anurag Srivastava 9

10 We draw an Amperian loop and the direction of integration. Amperian Loop Direction of Integration Wire surface B v ds v θ = 0 Recall, Therefore, N v B. ds = v µ 0 i n n= 1 ( ) I B cos θds = B ds = B 2 B 2 π r = + µ 0 µ 0I B = 2πr The equation derived earlier. ( πr) The positive sign for the current collaborates that the direction of B was correct. Anurag Srivastava 10

11 Example. Magnetic Field inside a Long Straight wire with current. R r B v ds v Wire surface Amperian Loop, B cos θds = B ds = B 2 N v v B. ds = µ 0 i n n= 1 ( πr), N v v B. ds = µ 0 i n n= 1 B cos θds B ds = B( 2πr = ) The charge enclosed is proportional to the area encircled by the loop, 2 πr i enc = i πr 2 Anurag Srivastava 11

12 The current enclosed is positive from the right hand rule. 2 πr B( 2π r) = µ 0i 2 πr µ 0i B = r 2 2πR Anurag Srivastava 12