Chapter 27 Magnetic Field an Magnetic Forces Study magnetic forces. Consider magnetic field and flux

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Chapter 27 Magnetic Field an Magnetic Forces Study magnetic forces Consider magnetic field and flux Explore motion in a magnetic field Consider magnetic torque Apply magnetic principles and study the electric motor Study the Hall effect Magnetic Forces only Act on Moving Charge Recall a charge produces and electric field A second charge responds to this field A moving charge produces and electric current An electric current produces a magnetic field A second electric current responds to the magnetic field A charge particle at rest experiences no magnetic force or magnetic field 1

Magnetism Magnetic north and south poles behavior is not unlike electric charges. For magnets, like poles repel and opposite poles attract. A permanent magnet will attract a metal like iron with either the north or south pole. The magnetic poles about our planet Magnetic poles reverse every 5000 to 50,000 yrs. Proof is from plate movement 2

Magnetic poles vs. Electric poles? We observed monopoles in electricity. A (+) or ( ) alone was stable and field lines could be drawn around it. Magnets cannot exist as monopoles. If you break a bar magnet between N and S poles, you get two smaller magnets, each with its own N and S pole. Electric current and magnets In 1820, Hans Oersted ran a series of experiments with conducting wires run near a sensitive compass. The result was dramatic. The orientation of the wire and the direction of the flow both moved the compass needle. There had to be something magnetic about current flow. 3

The interaction of magnetic force and charge The moving charge interacts with the fixed magnet. The force between them is at a maximum when the velocity of the charge is perpendicular to the magnetic field. Force = F = q v B B = Magnetic Field = F/qv = 1N-s/C (1A = 1C/s ) The units of B are: Tesla s T = 1N/A m 1T = 10,000 gauss (G) The earth s magnetic field B =~1G B is a vector and is defined as the direction the north pole of a compass needle will point. The right-hand rule I This is for a positive charge moving in a magnetic field. Place your hand out as if you were getting ready for a handshake. Your fingers represent the velocity vector of a moving charge. Move the fingers of your hand toward the magnetic field vector. Your thumb points in the direction of the force between the two vectors. 4

Right-hand rule II Two charges of equal magnitude but opposite signs moving in the same direction in the same field will experience force in opposing directions. Note: This relationship was not deduced theoretically; it is an observation based on experiment Cross Product Unit Vectors -j x k = -i i x k = -j -i x k = j j x k = i 5

Direction of a magnetic field with your CRT A TV or a computer screen is a cathode ray tube, an electron gun with computer aiming control. Place it in a magnetic field going up and down. You point the screen toward the ceiling and nothing happens to the picture. The magnetic field is parallel to the electron beam. You set the screen in a normal viewing position and the image distorts. The magnetic force is opposite to the thumb in the RHR. i x j =k -k for negative charge Force for a Negative charge Magnetic field lines may be traced Magnetic field lines may be traced from N toward S in analogous fashion to the electric field lines. 6

Magnetic Field Lines for common sources Magnetic flux through an area We define the magnetic flux through a surface just as we defined electric flux. Figure 27.15 illustrates the phenomenon. Follow Example 27.2, illustrated by Figure 27.16. 7

Motion of charged particles in a magnetic field A charged particle will move in a plane perpendicular to the magnetic field. Figure 27.17 at right illustrates the forces and shows an experimental example. Figure 27.18 below shows the constant kinetic energy and helical path. Test Your Understanding 8

Motion of Charged Particles in a Magnetic Field (Chapter 27, Sec 4) F q( vxb) ( vxb) vbsin sin sin90 1 2 v F qvbsin qvb m R 2 v qvb m v R R m q B Because F is always perpendicular to v, v is constant. Therefore, the charge will travel in a circle with radius R. 17 A magnetic bottle If we ever get seriously close to smalllab nuclear fusion, the magnetic bottle will likely be the only way to contain the unimaginable temperatures ~ 10 6 K. Figure 27.19 diagrams the magnetic bottle and Figure 27.20 shows the realworld examples northern lights and southern lights. 9

J.J. Thompson was able to characterize the electron Thompson s experiment was an exceptionally clever combination of known electron acceleration and magnetic steering. Electrons reach the screen for equal electric and magnetic forces Bainbridge s mass spectrometer Using the same concept as Thompson, Bainbridge was able to construct a device that would only allow one mass in flight to reach the detector. The fields could be ramped through an experiment containing standards (most high vacuum work always has a peak at 18 amu). -qe + qvb = 0 Solve for v: v = E / B R = mv / qb Larger mass larger radius 10

The magnetic force on a current-carrying conductor The force is always perpendicular to the conductor and the field. J x k = -i F = qv x B F = (nal) (qv d B) F = (nqv d ) (lb) J=nqv d I = JA = nqv d A F = I l B F = I l B sin a Magnetic Force on a straight Conductor 11

Straight Conductor Maximum Force If the weight of the rod is 60.0N and the mass 6.12 kg (m=60/9.8=6.12 kg) The rod will levitate: High-sped trains will levitate over the rails Magnetic force on a curved conductor -i x k = j F = IB (L + 2R)j Only the path normal to magnetic field determines the force 12

Force on a Conducting Rail F=I x B (-ji) x (-kb) =? or (+ji) x (-kb) =? Force and torque on a current loop This basis of electric motors is well diagrammed in Figure 27.31 below. -k x j = i 13

Force and Torque on a Current Loop F IlBsin90 F IaB b 2F sin Fb sin 2 IBab sin A = ab = area of coil For an N turn coil NIBA sin 27 Potential Energy of a Magnetic Dipole 14

The Direct-Current Motor Power for Electric Motor V ab = E + I r V ab = applied voltage E = back emf r = interal resistance of wire loop Page-942 29 The Hall Effect Potential proportional to magnetic field Considers the forces on charge carriers as they move through a conductor in a magnetic field. 15

Q27.1 When a charged particle moves through a magnetic field, the direction of the magnetic force on the particle at a certain point is A. in the direction of the magnetic field at that point. B. opposite to the direction of the magnetic field at that point. C. perpendicular to the magnetic field at that point. D. none of the above E. The answer depends on the sign of the particle s electric charge. Q27.2 A particle with a positive charge moves in the xz-plane as shown. The magnetic field is in the positive z-direction. The magnetic force on the particle is in A. the positive x-direction. B. the negative x-direction. C. the positive y-direction. D. the negative y-direction. E. none of these 16

Q27.3 A particle with charge q = 1 C is moving in the positive z-direction at 5 m/s. The magnetic field at its position is B = 3iˆ 4 ˆj T What is the magnetic force on the particle? ˆj ˆj 20iˆ+ 15 ˆj N 20iˆ 15 ˆj N 20iˆ+ 15 N 20iˆ15 N A. B. C. D. E. none of these Q27.4 A positively charged particle moves in the positive z-direction. The magnetic force on the particle is in the positive y-direction. What can you conclude about the x-component of the magnetic field at the particle s position? A. B x > 0 B. B x = 0 C. B x < 0 D. not enough information given to decide 17

Q27.5 A positively charged particle moves in the positive z-direction. The magnetic force on the particle is in the positive y-direction. What can you conclude about the y-component of the magnetic field at the particle s position? A. B y > 0 B. B y = 0 C. B y < 0 D. not enough information given to decide Q27.6 A positively charged particle moves in the positive z-direction. The magnetic force on the particle is in the positive y-direction. What can you conclude about the z-component of the magnetic field at the particle s position? A. B z > 0 B. B z = 0 C. B z < 0 D. not enough information given to decide 18

Q27.7 Under what circumstances is the total magnetic flux through a closed surface positive? A. if the surface encloses the north pole of a magnet, but not the south pole B. if the surface encloses the south pole of a magnet, but not the north pole C. if the surface encloses both the north and south poles of a magnet D. none of the above Q27.8 When a charged particle moves through a magnetic field, the trajectory of the particle at a given point is A. parallel to the magnetic field line that passes through that point. B. perpendicular to the magnetic field line that passes through that point. C. neither parallel nor perpendicular to the magnetic field line that passes through that point. D. any of the above, depending on circumstances 19

Q27.9 A charged particle moves through a region of space that has both a uniform electric field and a uniform magnetic field. In order for the particle to move through this region at a constant velocity, A. the electric and magnetic fields must point in the same direction. B. the electric and magnetic fields must point in opposite directions. C. the electric and magnetic fields must point in perpendicular directions. D. The answer depends on the sign of the particle s electric charge. Q27.10 A circular loop of wire carries a constant current. If the loop is placed in a region of uniform magnetic field, the net magnetic force on the loop is A. perpendicular to the plane of the loop, in a direction given by a right-hand rule. B. perpendicular to the plane of the loop, in a direction given by a left-hand rule. C. in the same plane as the loop. D. zero. E. The answer depends on the magnitude and direction of the current and on the magnitude and direction of the magnetic field. 20

Q27.11 A circular loop of wire carries a constant current. If the loop is placed in a region of uniform magnetic field, the net magnetic torque on the loop A. tends to orient the loop so that its plane is perpendicular to the direction of the magnetic field. B. tends to orient the loop so that its plane is edge-on to the direction of the magnetic field. C. tends to make the loop rotate around its axis. D. is zero. E. The answer depends on the magnitude and direction of the current and on the magnitude and direction of the magnetic field. 21