Today’s Concept: Ampere’s Law

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Presentation transcript:

Today’s Concept: Ampere’s Law Physics 2112 Unit 15 Today’s Concept: Ampere’s Law

Ampere’s Law We know for an infinite current carrying wire But what is 2pR? Circumference of circle!

Current enclosed by that closed loop Ampere’s Law Any closed loop Current enclosed by that closed loop

Checkpoint 1A Two loops are placed near identical current carrying wires as shown in Case 1 and Case 2 below. For which loop is ∫B·dl greater? Case 1 Case 2 The integral is the same for both

Checkpoint 1B Two loops are placed near identical current carrying wires as shown in Case 1 and Case 2 below. For which loop is ∫B·dl greater? Case 1 Case 2 The integral is the same for both

Checkpoint 1C Two loops are placed near current carrying wires as shown in Case 1 and Case 2 below. In both cases the direction of the current in the two wires are opposite to each other. For which loop is ∫B·dl greater? Case 1 Case 2 The integral is the same for both

Example 15.1 (B field from a thick wire) A wire with a radius of r=1cm has a uniform current of 1A flowing through it. What is the B field 1 meter from the center of the wire? What is the B field 0.5cm from the center of the wire?

Example 15.1 (graph) X X X X X X X X X A wire with a radius of r=1cm has a uniform current of 1A flowing through it. B r

CheckPoint 2A An infinitely long hollow conducting tube carries current I in the direction shown. X What is the direction of the magnetic field inside the tube? clockwise counterclockwise radially inward to the center radially outward from the center the magnetic field is zero

Why is that? X X

Example 15.3 (Pipe of current) y An infinitely long cylindrical shell carries a uniformly distributed current of 5A out of the screen. The inner radius is a=4cm and outer radius=8cm What is B at r = 2cm? What is B at r = 6cm? What is B at r = 16cm? I a x b The purpose of this Check is to jog the students minds back to when they studied work and potential energy in their intro mechanics class.

Example 15.3 (Pipe of current) Conceptual Analysis Complete cylindrical symmetry (can only depend on r)  can use Ampere’s law to calculate B Strategic Analysis Calculate B for the three regions separately: 1) r < a 2) a < r < b 3) r > b The purpose of this Check is to jog the students minds back to when they studied work and potential energy in their intro mechanics class. For circular path concentric with shell.

Example 15.2 (Two thick wires) Two thick cables both of radius R and length L carry a current, I, side-by-side as shown to the left. The left cable has the current into the screen and the right cable has the current out of the screen. A B X C D What is the magnetic field at points A, B, C and D? - A is at the center of the left cable - B is a distance R/2 from the center of the left cable - C is at the point where the cables touch. - D is the bottom of the left cable

The Plan What is the magnetic field at points A, B, C and D? X C D Conceptual Analysis use Ampere’s law to calculate B from each cable separately Strategic Analysis Note direction of B from each cable at each point. Add B like vectors

Sign of the dot product I into screen

Ampere’s Law dl B dl B dl B

Sign of the dot product dl B dl B dl B

Sign of the dot product dl B dl B dl B

B Field in Solenoid ~cancel out X X X X . . . . add up

. . . . . . . . B Field in Solenoid X X L n = turns per unit length (Ideal Solenoid L>>>r)

Solenoid   Bar Magnet

CheckPoint 2B A current carrying wire is wrapped around cardboard tube as shown below. In which direction does the magnetic field point inside the tube? left right up down out of the screen into the screen

Example 15.3 (B from Solenoid) A long thin solenoid consists of 500 turns of wire carrying 1.5A. It has a length of 250cm and a radius of 1cm. What is the magnetic field in the exact center of the coil? What is the contribution to this field from the 250th coil (the center coil)?

. . . . . . . . Make Sense? Note: B250/Btot = 0.024 (1/500 = 0.002) X X . . . . . . . . Note: B250/Btot = 0.024 (1/500 = 0.002) So closest coil contributes over 10 tens more than average coil. Make sense?

When does “Loop”  “Solenoid”?

A hint…..