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Lecture 16 Magnetic Fields& Force

Lecture 16 Magnetic Fields& Force. I. F. B. B. F. a/2. b. F. F. a. Torque on a Current Loop. Imagine a current loop in a magnetic field as follows:. In a motor, one has “N” loops of current. Example:. 30.0 o.

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Lecture 16 Magnetic Fields& Force

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  1. Lecture 16Magnetic Fields& Force General Physics II, Lec 16, By/ T.A. Eleyan

  2. I F B B F a/2 b F F a Torque on a Current Loop Imagine a current loop in a magnetic field as follows: In a motor, one has “N” loops of current General Physics II, Lec 16, By/ T.A. Eleyan

  3. Example: 30.0o A circular loop of radius 50.0 cm is oriented at an angle of 30.0o to a magnetic field of 0.50 T. The current in the loop is 2.0 A. Find the magnitude of the torque. B General Physics II, Lec 16, By/ T.A. Eleyan

  4. Galvanometer Device used in the construction of ammeters and voltmeters. General Physics II, Lec 16, By/ T.A. Eleyan

  5. Galvanometer used as Ammeter • Typical galvanometer have an internal resistance of the order of 60 W - that could significantly disturb (reduce) a current measurement. • Built to have full scale for small current ~ 1 mA or less. • Must therefore be mounted in parallel with a small resistor or shunt resistor. 60 W Galvanometer Rp General Physics II, Lec 16, By/ T.A. Eleyan

  6. Galvanometer used as Voltmeter • Finite internal resistance of a galvanometer must also addressed if one wishes to use it as voltmeter. • Must mounted a large resistor in series to limit the current going though the voltmeter to 1 mA. • Must also have a large resistance to avoid disturbing circuit when measured in parallel. 60 W Galvanometer Rs General Physics II, Lec 16, By/ T.A. Eleyan

  7. Motion of Charged Particle in magnetic field • Consider positively charge particle moving in a uniform magnetic field. • Suppose the initial velocity of the particle is perpendicular to the direction of the field. • Then a magnetic force will be exerted on the particle and make follow a circular path. General Physics II, Lec 16, By/ T.A. Eleyan

  8. The magnetic force produces a centripetal acceleration. The particle travels on a circular trajectory with a radius: What is the period of revolution of the motion? The frequency General Physics II, Lec 16, By/ T.A. Eleyan

  9. Example :Proton moving in uniform magnetic field A proton is moving in a circular orbit of radius 14 cm in a uniform magnetic field of magnitude 0.35 T, directed perpendicular to the velocity of the proton. Find the orbital speed of the proton. General Physics II, Lec 16, By/ T.A. Eleyan

  10. v x x r x x Example: If a proton moves in a circle of radius 21 cm perpendicular to a B field of 0.4 T, what is the speed of the proton and the frequency of motion? General Physics II, Lec 16, By/ T.A. Eleyan

  11. [Q]: Consider the mass spectrometer. The electric field between the plates of the velocity selector is 950 V/m, and the magnetic fields in both the velocity selector and the deflection chamber have magnitudes of 0.930 T. Calculate the radius of the path in the system for a singly charged ion with mass m=2.18×10-26 kg. General Physics II, Lec 16, By/ T.A. Eleyan

  12. Example: Mass Spectrometer • Suppose that B=80mT, V=1000V. A charged ion (1.6022 10-19C) enters the chamber and strikes the detector at a distance x=1.6254m. What is the mass of the ion? • Key Idea: The uniform magnetic field forces the ion on a circular path and the ion’s mass can be related to the radius of the circular trajectory. General Physics II, Lec 16, By/ T.A. Eleyan

  13. General Physics II, Lec 16, By/ T.A. Eleyan

  14. Cyclotrons • A cyclotron is a particle accelerator • The D-shaped pieces (descriptively called “dees”) have alternating electric potentials applied to them such that a positively charged particle always sees a negatively charged dee ahead when it emerges from under the previous dee, which is now positively charged General Physics II, Lec 16, By/ T.A. Eleyan

  15. The resulting electric field accelerates the particle • Because the cyclotron sits in a strong magnetic field, the trajectory is curved • The radius of the trajectory is proportional to the momentum, so the accelerated particle spirals outward General Physics II, Lec 16, By/ T.A. Eleyan

  16. Example: Deuteron in Cyclotron Suppose a cyclotron is operated at frequency f=12 MHz and has a dee radius of R=53cm. What is the magnitude of the magnetic field needed for deuterons to be accelerated in the cyclotron (m=3.34 10-27kg)? General Physics II, Lec 16, By/ T.A. Eleyan

  17. Example Suppose a cyclotron is operated at frequency f=12 MHz and has a dee radius of R=53cm. What is the kinetic energy of the deuterons in this cyclotron when they travel on a circular trajectory with radius R (m=3.34 10-27kg, B=1.57 T)? A) 0.9 10-14 J B) 8.47 10-13 J C) 2.7 10-12 J D) 3.74 10-13 J General Physics II, Lec 16, By/ T.A. Eleyan

  18. Lecture 17Sources of the Magnetic Field General Physics II, Lec 16, By/ T.A. Eleyan

  19. History • 1819 Hans Christian Oersted discovered that a compass needle was deflected by a current carrying wire • Then in 1920s Jean-Baptiste Biot and Felix Savart performed experiments to determine the force exerted on a compass by a current carrying wire General Physics II, Lec 16, By/ T.A. Eleyan

  20. Biot & Savart’s Results • dB the magnetic field produced by a small section of wire • ds a vector the length of the small section of wire in the direction of the current • r the positional vector from the section of wire to where the magnetic field is measured • I the current in the wire • angle between ds & r General Physics II, Lec 16, By/ T.A. Eleyan

  21. dB perpendicular to ds • |dB| inversely proportional to |r|2 • |dB| proportional to current I • |dB| proportional to |ds| • |dB| proportional to sin q General Physics II, Lec 16, By/ T.A. Eleyan

  22. Biot–Savart Law All these results could be summarised by one “Law” Putting in the constant Where m0 is the permeablity of free space General Physics II, Lec 16, By/ T.A. Eleyan

  23. Magnetic Field due to Currents The passage of a steady current in a wire produces a magnetic field around the wire. • Field form concentric lines around the wire • Direction of the field given by the right hand rule. If the wire is grasped in the right hand with the thumb in the direction of the current, the fingers will curl in the direction of the field. • Magnitude of the field General Physics II, Lec 16, By/ T.A. Eleyan

  24. [Q]: The two wires in the figure below carry currents of 3.00A and 5.00A in the direction indicated. Find the direction and magnitude of the magnetic field at a point midway between the wires. 5.00 A 3.00 A 20.0 cm General Physics II, Lec 16, By/ T.A. Eleyan

  25. Magnetic Field of a current loop Dx1 Magnetic field produced by a wire can be enhanced by having the wire in a loop. 1 loop Current I B I N loops Current  NI Dx2 General Physics II, Lec 16, By/ T.A. Eleyan

  26. [Q] What is the magnetic field at point Q in Fig.? [Q] What is the magnitude and direction of the magnetic field at point P in Fig.? General Physics II, Lec 16, By/ T.A. Eleyan

  27. [Q] Use the Biot-Savart Law to calculate the magnetic field B at C, the common center of the semicircular area AD and HJ of radii R1=8 cm and R2=4 cm, forming part of the circuit ADJHA carrying current I=10 A, as seen figure. General Physics II, Lec 16, By/ T.A. Eleyan

  28. Ampere’s Law Consider a circular path surrounding a current, divided in segments ds, Ampere showed that the sum of the products of the field by the length of the segment is equal to mo times the current. General Physics II, Lec 16, By/ T.A. Eleyan

  29. Example By way of illustration, let us use Ampere's law to find the magnetic field at a distance r from a long straight wire, a problem we have solved already using the Biot-Savart law General Physics II, Lec 16, By/ T.A. Eleyan

  30. Magnetic Force between two parallel conductors General Physics II, Lec 16, By/ T.A. Eleyan

  31. Force per unit length General Physics II, Lec 16, By/ T.A. Eleyan

  32. Definition of the SI unit Ampere Used to define the SI unit of current called Ampere. If two long, parallel wires 1 m apart carry the same current, and the magnetic force per unit length on each wire is 2x10-7 N/m, then the current is defined to be 1 A. General Physics II, Lec 16, By/ T.A. Eleyan

  33. Example Two wires, each having a weight per units length of 1.0x10-4 N/m, are strung parallel to one another above the surface of the Earth, one directly above the other. The wires are aligned north-south. When their distance of separation is 0.10 mm what must be the current in each in order for the lower wire to levitate the upper wire. (Assume the two wires carry the same current). l 1 I1 2 d I2 General Physics II, Lec 16, By/ T.A. Eleyan

  34. F1 1 I1 B2 mg/l 2 d I2 l General Physics II, Lec 16, By/ T.A. Eleyan

  35. Magnetic Field of a solenoid • Solenoid magnet consists of a wire coil with multiple loops. • It is often called an electromagnet. General Physics II, Lec 16, By/ T.A. Eleyan

  36. Solenoid Magnet • Field lines inside a solenoid magnet are parallel, uniformly spaced and close together. • The field inside is uniform and strong. • The field outside is non uniform and much weaker. • One end of the solenoid acts as a north pole, the other as a south pole. • For a long and tightly looped solenoid, the field inside has a value: General Physics II, Lec 16, By/ T.A. Eleyan

  37. Since the field lines are straight inside the solenoid, the best choice for amperian loop is a rectangle: abcd. Winding density: n=N/L where N = total number of windings and L = total length. Integrate: General Physics II, Lec 16, By/ T.A. Eleyan

  38. Solenoid Magnet The field inside has a value: n = N/L : number of (loop) turns per unit length. I : current in the solenoid. General Physics II, Lec 16, By/ T.A. Eleyan

  39. Example: Consider a solenoid consisting of 100 turns of wire and length of 10.0 cm. Find the magnetic field inside when it carries a current of 0.500 A. General Physics II, Lec 16, By/ T.A. Eleyan

  40. General Physics II, Lec 16, By/ T.A. Eleyan

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