2. Physics 1202: Lecture 11 Today’s Agenda Announcements: –Lectures posted on: www.phys.uconn.edu/~rcote/ www.phys.uconn.edu/~rcote/ –HW assignments, solutions etc. Homework #4:Homework #4: –Not this week ! (time to prepare midterm) Midterm 1: –Friday Oct. 2 –Chaps. 15, 16 & 17.

4. Lorentz Force The force F on a charge q moving with velocity v through a region of space with electric field E and magnetic field B is given by: F x x x v B q  v B q F = 0  v B q F  Units: 1 T (tesla) = 1 N / Am 1G (gauss) = 10 -4 T

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. Magnetic Force on a Current Consider a current-carrying wire in the presence of a magnetic field B. There will be a force on each of the charges moving in the wire. What will be the total force  F on a length  l of the wire? Suppose current is made up of n charges/volume each carrying charge q and moving with velocity v through a wire of cross- section A. N S Simpler: For a straight length of wire L carrying a current I, the force on it is: Force on each charge = Total force =  Current = or

9. Lecture 11, ACT1 A current I flows in a wire which is formed in the shape of an isosceles triangle as shown. A constant magnetic field exists in the -z direction. –What is F y, net force on the wire in the y- direction? (a) F y < 0 (b) F y = 0 (c) F y > 0 x y

10. Magnetic Force on a Current Loop 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 x x x x x x x B i Consider loop in magnetic field as on right: If field is  to plane of loop, the net force on loop is 0! If plane of loop is not  to field, there will be a non-zero torque on the loop! B x. F F F F –Force on top path cancels force on bottom path ( F = IBL ) F F –Force on right path cancels force on left path. ( F = IBL )

11. Calculation of Torque Suppose a square wire loop has width w (the side we see) and length L (into the screen). The torque is given by: rxF Note: if loop  B, sin  = 0   = 0 maximum  occurs when loop parallel to B since: A = wL = area of loop r F  B x. F F  w  

12. Magnetic Dipole Moment We can define the magnetic dipole moment of a current loop as follows: direction:  to plane of the loop in the direction the thumb of right hand points if fingers curl in direction of current. Torque on loop can then be rewritten as: Note: if loop consists of N turns,  = N A I magnitude:  A I  A I B  sin  B x. F F  w   

13. Electric Dipole Analogy E. +q -q p F F B x. F F w   (per turn)

14. Lecture 11, ACT 2 A rectangular loop is placed in a uniform magnetic field with the plane of the loop parallel to the direction of the field. If a current is made to flow through the loop in the sense shown by the arrows, the field exerts on the loop: A) a net force. B) a net torque. C) a net force and a net torque. D) neither a net force nor a net torque. Solution 1.Bottom part force is out of page. Sides force = 0. Top force into page. 2.Sum of all forces is zero. 3.But torques,  = r x F, not zero, I.e. loop will rotate. 4.Answer is B.

15. Lecture 11, ACT 3 A circular loop of radius R carries current I as shown in the diagram. A constant magnetic field B exists in the +x direction. Initially the loop is in the x-y plane. –The coil will rotate to which of the following positions? (a) (b) (c) It will not rotate

16. 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"  0 = 4  X 10 -7 T m /A: permeability (vacuum)

17. Magnetic Field of  Straight Wire  Direction of B: right-hand rule

18. Lecture 11, ACT 3 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 By right hand rule, B at wire 2 due to wire 1 is down. B The current direction is into the page. From last lectures, F = I L x B, cross product to left F

19. 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

20. Circular Loop x z R R Circular loop of radius R carries current i. Calculate B along the axis of the loop: r BB r z BB   Symmetry  B in z-direction. > > I  At the center (z=0): z>>R: Note the form the field takes for z>>R: for N coils

21. Lecture 11, ACT 4 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 The right current loop gives rise to B z <0 at point A. The left current loop gives rise to B z >0 at point A. From symmetry, the magnitudes of the fields must be equal. Therefore, B(A) = 0!

22. 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? The signs of the fields from each loop are the same at B as they are at A! However, point B is closer to the right loop, so its field wins!

24. 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

27. 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). (n: number of turns per unit length) 

29. Toroid Toroid defined by N total turns with current i. B=0 outside toroid! B inside the toroid. x x x x x x x x x x x x x x x x r B 

30. 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