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Physics 1502: Lecture 17 Today’s Agenda Announcements: –Midterm 1 distributed today Homework 05 due FridayHomework 05 due Friday Magnetism.

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Presentation on theme: "Physics 1502: Lecture 17 Today’s Agenda Announcements: –Midterm 1 distributed today Homework 05 due FridayHomework 05 due Friday Magnetism."— Presentation transcript:

1 Physics 1502: Lecture 17 Today’s Agenda Announcements: –Midterm 1 distributed today Homework 05 due FridayHomework 05 due Friday Magnetism

2 Trajectory in Constant B Field F F v R x x x v B q Suppose charge q enters B field with velocity v as shown below. (v  B) What will be the path q follows? Force is always  to velocity and B. What is path? –Path will be circle. F will be the centripetal force needed to keep the charge in its circular orbit. Calculate R:

3 Radius of Circular Orbit Lorentz force: centripetal acc: Newton's 2nd Law: x x x v F B q F v R This is an important result, with useful experimental consequences !  

4 Ratio of charge to mass for an electron e-e- 3) Calculate B … next week; for now consider it a measurement 4) Rearrange in terms of measured values, V, R and B 1) Turn on electron ‘gun’ VV ‘gun’ 2) Turn on magnetic field B R & 

5 Lawrence's Insight "R cancels R" We just derived the radius of curvature of the trajectory of a charged particle in a constant magnetic field. E.O. Lawrence realized in 1929 an important feature of this equation which became the basis for his invention of the cyclotron. R does indeed cancel R in above eqn. So What?? –The angular velocity is independent of R!! –Therefore the time for one revolution is independent of the particle's energy! –We can write for the period, T=2  /  or T = 2  m/qB –This is the basis for building a cyclotron. Rewrite in terms of angular velocity  !   

6 The Hall Effect c d l a c B B I I - vdvd F Hall voltage generated across the conductor qE H Force balance Using the relation between drift velocity and current we can write:

7 The Laws of Biot-Savart & Ampere  x R r   P I dx dl I

8 Calculation of Electric Field Two ways to calculate the Electric Field: Coulomb's Law: Gauss' Law What are the analogous equations for the Magnetic Field? "Brute force" "High symmetry"

9 Calculation of Magnetic Field Two ways to calculate the Magnetic Field: Biot-Savart Law: Ampere's Law These are the analogous equations for the Magnetic Field! "Brute force"  I "High symmetry"

10 Biot-Savart Law… bits and pieces I dl dB X r  So, the magnetic field “circulates” around the wire B in units of Tesla (T) A  0 = 4  X 10 -7 T m /A

11 Magnetic Field of  Straight Wire Calculate field at point P using Biot-Savart Law: Rewrite in terms of R,  : x R r   P I dx Which way is B?   

12 Magnetic Field of  Straight Wire x R r   P I dx  

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14 Lecture 17, ACT 1 I have two wires, labeled 1 and 2, carrying equal current, into the page. We know that wire 1 produces a magnetic field, and that wire 2 has moving charges. What is the force on wire 2 from wire 1 ? (a) Force to the right (b) Force to the left (c) Force = 0 Wire 1 I X Wire 2 I X B F

15 Force between two conductors Force on wire 2 due to B at wire 1: Total force between wires 1 and 2: Force on wire 2 due to B at wire 1: Direction: attractive for I 1, I 2 same direction repulsive for I 1, I 2 opposite direction

16 Circular Loop x z R R Circular loop of radius R carries current i. Calculate B along the axis of the loop: Magnitude of dB from element dl: r dB r z   What is the direction of the field? Symmetry  B in z-direction. 

17 Circular Loop  Note the form the field takes for z>>R: Expressed in terms of the magnetic moment:  note the typical dipole field behavior! x z R R r r dB z  

18 Circular Loop R B z z 0 0  1 z 3

19 Lecture 17, ACT 2 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 B z (A) at point A, the midpoint between the two loops? (a) B z (A) < 0 (b) B z (A) = 0 (c) B z (A) > 0

20 Lecture 17, ACT 3 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. (a) B z (B) < 0 (b) B z (B) = 0 (c) B z (B) > 0 – What is the magnetic field B z (B) at point B, just to the right of the right loop?

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

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

25 Review: B Field of a Long Wire Inside the wire: (r < a) Outside the wire: (r>a) r B a

26 Lecture 17, ACT 4 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 x (a) at point a, just outside the cylinder as shown? 2A (a) B x (a) < 0 (b) B x (a) = 0 (c) B x (a) > 0

27 Lecture 17, ACT 4 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. 2B (a) B x (b) < 0 (b) B x (b) = 0 (c) B x (b) > 0 – What is the magnetic field B x (b) at point b, just inside the cylinder as shown?

28 B 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. If a << L, the B field is to first order contained within the solenoid, in the axial direction, and of constant magnitude. In this limit, we can calculate the field using Ampere's Law. L 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. a

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

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

34 Magnetic Flux Define the flux of the magnetic field through a surface (closed or open) from: Gauss’s Law in Magnetism dS B B

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36 Magnetism in Matter When a substance is placed in an external magnetic field B o, the total magnetic field B is a combination of B o and field due to magnetic moments (Magnetization; M): – B = B o +  o M =  o (H +M) =  o (H +  H) =  o (1+  ) H »where H is magnetic field strength  is magnetic susceptibility Alternatively, total magnetic field B can be expressed as : –B =  m H »where  m is magnetic permeability »  m =  o (1 +  ) All the matter can be classified in terms of their response to applied magnetic field: –Paramagnets  m >  o –Diamagnets  m <  o –Ferromagnets  m >>>  o


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