Physics 1402: Lecture 18 Today s Agenda

Transcription

1 Physics 1402: Lecture 18 Today s Agenda Announcements: Midterm 1 distributed available Homework 05 due Friday Magnetism Calculation of Magnetic Field Two ways to calculate the Magnetic Field: iot-savart Law: I "rute force" Ampere's Law "High symmetry" These are the analogous equations for the Magnetic Field! 1

2 Infinite line Magnetic Fields P θ r R Circular loop θ d I R θ z R r r d θ d z 2

3 Force between two conductors Force on wire 2 due to at wire 1: Force on wire 2 due to at wire 1: Total force between wires 1 and 2: Direction: attractive for I 1, I 2 same direction repulsive for I 1, I 2 opposite direction Lecture 18, ACT 1 Equal currents I flow in identical circular loops as shown in the diagram. The loop on the right (left) carries current in the ccw (cw) direction as seen looking along the +z direction. What is the magnetic field z (A) at point A, the midpoint between the two loops? (a) z (A) < 0 (b) z (A) = 0 (c) z (A) > 0 3

4 Lecture 18, ACT 1 Equal currents I flow in identical circular loops as shown in the diagram. The loop on the right (left) carries current in the ccw (cw) direction as seen looking along the +z direction. What is the magnetic field z () at point, just to the right of the right loop? (a) z () < 0 (b) z () = 0 (c) z () > 0 4

5 Magnetic Field of Straight Wire Calculate field at distance R from wire using Ampere's Law: Choose loop to be circle of radius R centered on the wire in a plane to wire. Why?» Magnitude of is constant (fct of R only)» Direction of is parallel to the path. Evaluate line integral in Ampere s Law: Current enclosed by path = I Apply Ampere s Law: I dl R Ampere's Law simplifies the calculation thanks to symmetry of the current! ( aial/cylindrical ) 5

6 Field inside a Long Wire? What is the field at a distance R, with R<a (a: radius of wire)? Radius a Choose loop to be circle of radius R, whose edges are inside the wire. I R Why?» Left Hand Side is same as before. Current enclosed by path = J Area of Loop Apply Ampere s Law: Inside the wire: (r < a) Review: Field of a Long Wire a Outside the wire: (r>a) r 6

7 Lecture 18, ACT 3 A current I flows in an infinite straight wire in the +z direction as shown. A 2A concentric infinite cylinder of radius R carries current I in the -z direction. What is the magnetic field (a) at point a, just outside the cylinder as shown? (a) (a) < 0 (b) (a) = 0 (c) (a) > 0 2 Lecture 18, ACT 3 A current I flows in an infinite straight wire in the +z direction as shown. A concentric infinite cylinder of radius R carries current I in the -z direction. What is the magnetic field (b) at point b, just inside the cylinder as shown? (a) (b) < 0 (b) (b) = 0 (c) (b) > 0 7

8 Field of a Solenoid A constant magnetic field can (in principle) be produced by an sheet of current. In practice, however, a constant magnetic field is often produced by a solenoid. A solenoid is defined by a current I flowing through a wire which is wrapped n turns per unit length on a cylinder of radius a and length L. L a If a << L, the field is to first order contained within the solenoid, in the aial direction, and of constant magnitude. In this limit, we can calculate the field using Ampere's Law. 8

9 Field of a Solenoid To calculate the field of the solenoid using Ampere's Law, we need to justify the claim that the field is 0 outside the solenoid. To do this, view the solenoid from the side as 2 current sheets. The fields are in the same direction in the region between the sheets (inside the solenoid) and cancel outside the sheets (outside the solenoid). Draw square path of side w: (n: number of turns per unit length) 9

10 Toroid Toroid defined by N total turns with current i. =0 outside toroid! (Consider integrating on circle outside toroid) To find inside, consider circle of radius r, centered at the center of the toroid. r Apply Ampere s Law: 10

11 Magnetic Flu Define the flu of the magnetic field through a surface (closed or open) from: ds Gauss s Law in Magnetism 11

12 Magnetism in Matter When a substance is placed in an eternal magnetic field o, the total magnetic field is a combination of o and field due to magnetic moments (Magnetization; M): = o + µ o M = µ o (H +M) = µ o (H + χ H) = µ o (1+χ) H» where H is magnetic field strength χ is magnetic susceptibility Alternatively, total magnetic field can be epressed as: = µ m H» where µ m is magnetic permeability» µ m = µ o (1 + χ ) All the matter can be classified in terms of their response to applied magnetic field: Paramagnets µ m > µ o Diamagnets µ m < µ o Ferromagnets µ m >>> µ o Faraday's Law ds N S v S N v 12

13 Induction Effects ar magnet moves through coil Current induced in coil S N v Change pole that enters Induced current changes sign N S v ar magnet stationary inside coil No current induced in coil N S Coil moves past fied bar magnet Current induced in coil v S N Faraday's Law Define the flu of the magnetic field through a surface (closed or open) from: ds Faraday's Law: The emf induced around a closed circuit is determined by the time rate of change of the magnetic flu through that circuit. The minus sign indicates direction of induced current (given by Lenz's Law). 13

14 Faraday s law for many loops Circuit consists of N loops: all same area Φ magn. flu through one loop loops in series emfs add! Lenz's Law: Lenz's Law The induced current will appear in such a direction that it opposes the change in flu that produced it. S N v N S v Conservation of energy considerations: Claim: Direction of induced current must be so as to oppose the change; otherwise conservation of energy would be violated.» Why??? If current reinforced the change, then the change would get bigger and that would in turn induce a larger current which would increase the change, etc.. 14

15 Lecture 18, ACT 4 y A conducting rectangular loop moves with constant velocity v in the + direction through a region of constant magnetic field in the -z direction as shown. What is the direction of the induced current in the loop? 4A (a) ccw (b) cw (c) no induced current Lecture 18, ACT 4 A conducting rectangular loop moves with constant velocity v in the -y direction away from a wire with a constant current I as shown. What is the direction of the 4 induced current in the loop? y i (a) ccw (b) cw (c) no induced current 15

16 Calculation Suppose we pull with velocity v a coil of resistance R through a region of constant magnetic field. What will be the induced current?» What direction? Lenz Law clockwise!! What is the magnitude? I w v» Magnetic Flu:» Faraday s Law: Δ E Faraday's law a changing induces an emf which can produce a current in a loop. In order for charges to move (i.e., the current) there must be an electric field. we can state Faraday's law more generally in terms of the E field which is produced by a changing field. E E r E E Suppose is increasing into the screen as shown above. An E field is induced in the direction shown. To move a charge q around the circle would require an amount of work = This work can also be calculated from 16

17 Δ E Putting these 2 eqns together: Therefore, Faraday's law can be rewritten in terms of the fields as: E E r E E Note that for E fields generated by charges at rest (electrostatics) since this would correspond to the potential difference between a point and itself. Consequently, there can be no "potential function" corresponding to these induced E fields. Lecture 18, ACT 5 The magnetic field in a region of space of radius 2R is aligned with the z-direction and 5A changes in time as shown in the plot. What is sign of the induced emf in a ring of radius R at time t=t 1? (a) ε < 0 ( E ccw) (b) ε = 0 (c) ε > 0 ( E cw) X X X X X X X X X X X X X X X X X X X X X X X X X X X X R X X X X X X X X X X X X X X X X X X X X X X X X X X X X z y t 1 t 17

18 5 Lecture 18, ACT 5 What is the relation between the magnitudes of the induced electric fields E R at radius R and E 2R at radius 2R? X X X X X X X X X X X X X X X X X X X X X X X X X X X X R X X X X X X X X X X X X X X X X X X X X X X X X X X X X z y (a) E 2R = E R (b) E 2R = 2E R (c) E 2R = 4E R t 1 t Eample (a) (b) An instrument based on induced emf has been used to measure projectile speeds up to 6 km/s. A small magnet is imbedded in the projectile, as shown in Figure below. The projectile passes through two coils separated by a distance d. As the projectile passes through each coil a pulse of emf is induced in the coil. The time interval between pulses can be measured accurately with an oscilloscope, and thus the speed can be determined. Sketch a graph of ΔV versus t for the arrangement shown. Consider a current that flows counterclockwise as viewed from the starting point of the projectile as positive. On your graph, indicate which pulse is from coil 1 and which is from coil 2. If the pulse separation is 2.40 ms and d = 1.50 m, what is the projectile speed? 18

19 A Loop Moving Through a Magnetic Field ε(t) =? F(t) =? Φ(t) =? Schematic Diagram of an AC Generator ε = Ν d Φ dt d (cos( ωt)) = ΝΑΒ dt = ΝΑΒ ω sin( ωt)) 19

20 Schematic Diagram of an DC Generator 20