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Magnetism The Magnetic Force r r r r F = qE + qv × B B x x x x x x x x x x x x v x x x x x x q F B →→→→→ v →→→→→ ´ q F B ↑↑↑↑↑↑↑↑ v ↑↑↑↑↑↑↑↑ q F=0 Magnetic Fields (History) • Earliest mention (~800 BC) of magnetic effects are from rocks, called lodestones, which are natural occurring permanent magnets. • Early seafarers used lodestones as compasses. The word, lode, means “way or course” in middle english, so lode+stone meant a “direction stone” for sailors. • Lodestones contain magnetite. It is believed that iron ore becomes magnetized when lightning struck the iron ore with high current and magnetized the iron, creating magnetite. There are many interesting legends about lodestones Magnetic Properties • All magnets are found to have a PAIR of points of attraction/repulsion, called the NORTH and SOUTH poles. • N & S poles attract • N & N or S & S poles repel • If the magnet is broken into two pieces, there are still a pair of North & South poles. There are no magnetic monopoles !!!! (none found so far). We live on a large magnet: planet Earth • Earth has a magnetic field indicated by the blue lines. • Compass, or free floating magnet, aligns its N-pole direction along the earth’s magnetic field lines. • Note the Earth’s magnetic South-pole is at the Earth’s geographic North pole Bearth~10-4T Bar magnet sprinkled with iron filings • Place a bar magnet underneath paper and then sprinkle iron filings on the paper, the filings behave like tiny compasses and align themselves along magnetic field lines. Magnetic Field lines are similar in direction to Electric Field lines from an electric dipole Sources of Magnetic Fields, B • In Chap 28 we will study the sources of magnetic fields, which are created from permanent magnets, currents and moving charges • For now (this week in Chap 27), let’s assume there exist vector magnetic fields. They will look similar to electric fields, that is at each position in space there are a field lines with direction and magnitude. The unit of the magnetic field is given in units of Tesla (“T”). We will mostly be concerned with constant magnetic field examples. Lorentz Force Law • The force on a moving charged particle is given by, r r r F = qv × B where q is the charge, v is the vector velocity and B is the magnetic Field in units of Tesla, T. • NOTE if the particle is not moving, there is NO force from B field • NOTE the force is perpendicular to the velocity, hence the magnitude of the velocity does not change, but the direction does. This is similar to centripetal acceleration studied in Phys 170. If an E field is also present then r r r r F = qE + qv × B Vector Cross Product r r r C = A× B Use right hand rule to get vector C direction and ABsinθ to get magnitude r r r C = A B sin θ r C θ r A r B Rotate A into B with your finger tips. Thumb points into C direction 2nd Useful formula for vector cross product in component form r C = (Ay Bz − Az B y )xˆ + ( Az Bx − Ax Bz ) yˆ + (Ax B y − Ay Bx )zˆ Divergence and Curl in spherical coordinates More on the cross product ) ) ) Unit vectors: i ×k = − j ) ) ) Check with right i× j =k hand rule ) ) ) j ×k = i ) ) ) ) ) ) i × i = 0; j × j = 0; k × k = 0 Example: A particle with q = -1.24 x 10-8 C and velocity ) ) 4 v = (4.19 ×10 m / sr)i + (−3.85 × 10 m / s ) j ) enters a region of space with B = (1.40T )i 4 Find the force on it: r ) ) ) ) r v F = qv × B = (−1.24 ×10−8 C )(1.40T )[(4.19 ×104 m / s )i × i − (3.85 ×104 m / s ) j × i ] r ) ) r v −8 4 F = qv × B = (−1.24 × 10 C )(1.40T )[(3.85 ×10 m / s ) j × i ] r ) r v −4 F = qv × B = 6.68 ×10 N (−k ) Clicker: Three points are arranged in a uniform magnetic field. The B field points into the screen. r r r Magnetic Force: F = qv × B 1) A positively charged particle is located at point A and is stationary. The direction of the magnetic force on the particle is: a) right b) left d) out of the screen c) into the screen If v = 0 ⇒ F = 0. e) zero 2) The positive charge moves from point A toward B. The direction of the magnetic force on the particle is: r r If a) right b) left d) out of the screen v⊥B then F = qvB c) into the screen e) zero If v is up, and B is into the page, then F is to the left. Clicker question: 3) The positive charge moves from point A toward C. The direction of the magnetic force on the particle is: a) up and right b) up and left c) down and right d) down and left r r r Magnetic Force: F = q v × B If v is up and to the right, it is still perpendicular to B, hence F = qvB then and F is up and to the left. Clicker problem • Two protons each move at speed v (as shown in the diagram) in a region of space which contains a constant B field 2A in the -z-direction. Ignore the interaction between the two protons. – What is the relation between the magnitudes of the forces on the two protons? (a) F1 < F2 2B 1 v B 2 z v x (c) F1 > F2 – What is F2x, the x-component of the force on the second proton? (a) F2x < 0 2C (b) F1 = F2 y (b) F2x = 0 (c) F2x > 0 – Inside the B field, the speed of each proton: (a) decreases (b) increases (c) stays the same Clicker problem • Two independent protons each move at speed v (as shown in the diagram) in a region of space which contains a 2A constant B field in the -z-direction. Ignore the interaction between the two protons. – What is the relation between the magnitudes of the forces on the two protons? (a) F1 < F2 (b) F1 = F2 y 1 v B 2 v z (c) F1 > F2 • The magnetic force is given by: r r r F = qv ´ B ⇒ F = qvB sin θ • In both cases the angle between v and B is 90°!! Therefore F1 = F2. x Clicker problem • Two independent protons each move at speed v (as shown in the diagram) in a region of space which contains a 2B constant B field in the -z-direction. Ignore the interaction between the two protons. – What is F2x, the x-component of the force on the second proton? (a) F2x < 0 (b) F2x = 0 F1 F2 y 1 v B 2 v z x (c) F2x > 0 • To determine the direction of the force, we use the right-hand rule. r r r F = qv × B • As shown in the diagram, F2x < 0. Clicker problem • Two protons each move at speed v (as 2C shown in the diagram) in a region of space which contains a constant B field in the -z-direction. Ignore the interaction between the two protons. – Inside the B field, the speed of each proton: (a) decreases (b) increases y 1 v B 2 z v x (c) stays the same Although the proton does experience arforce (which deflects it), this is always ⊥ to v . Therefore, there is no possibility to do work, so r kinetic energy is constant and v is constant. Magnetic Flux • Recall that we introduced the electric flux r r Φ E = ∫ E ⋅ dA and used it in Gauss’ Law E dA S r r qin Φ E = ∫ E ⋅ dA = S ε0 • Lets introduce the analog, the magnetic flux as the surface integral of the normal component of the magnetic rfield r Φ B = ∫ B ⋅ dA S The units are Tesla·m2, which are called Weber (Wb)’s Magnetic Flux Law There are no magnetic monopoles ! • We now state the magnetic flux law (analogous to Gauss’ Law) r r Φ M = ∫ B ⋅ dA = 0 • The net magnetic flux through any enclosed surface is always ZERO This is equivalent to the non-existence of magnetic monopoles and the fact that all magnetic lines form a loop (do not start/end on a charge). S S Any enclosed surface will have net magnetic flux = zero above are B fields from bar magnet and current loop Example (not a clicker problem) The magnetic flux through one face of a cube is +120Wb A) What must the total magnetic flux through the other five faces of the cube be? B) Why did you not need to know the dimensions of the cube in order to answer part A)? C) Suppose the magnetic flux is due to a permanent magnet, show Where the cube in part (A) might be be located A) -120 Wb ?? +120 Wb B) Flux is zero for a closed surface C) Just outside the block, with equal # lines leaving/entering Comparison between bar magnet & E-dipole The magnetic lines form closed loops. The electric fields start on the +charges and end on the –charges. So if we place a closed Gaussian box around a charge, we find non-zero electric flux. If we place a closed box anywhere in the bar magnet case, we always obtain zero magnetic flux. There are NO free “magnetic charges”. These hypothetical particles are called “magnetic monopoles”. r r Q ∫ E ⋅ dA = ε0 r r ∫ B ⋅ dA = 0 The above two equations are TWO of the FOUR fundamental equations from Maxwell, called Maxwell’s Eqns. (see Y&F pg.1216 for the other two). Maxwell’s Eqns. predict and govern all of the physics of electricity and magnetism. Congratulations on understanding HALF of all electromagnetic science. Charged particle motion in a constant B field - velocity in plane to B. r Suppose we have a magnetic field given by B = − B0 zˆ and a particle starts out at the origin moving in the +x direction. The particle will move in a circle with the radius R and angular velocity ω: r r v2 F = m a = qvB = m R y mv R= qB θ magnetic force provides centripedal force R v qB ω= = R m v x independent of r and v. Review of Circular Motion If the force F is perpendicular to the velocity, the particle undergoes circular motion: r r mv 2 F = ma = R v R F Force F is always perpendicular to velocity. F does no work. Example from Y+F In an experiment with cosmic rays, a vertical beam of particles that have a charge magnitude of 3e and a mass 12 times the proton mass enters a uniform horizontal Magnetic field of 0.250T and is bent in a semicircle of 95cm. Find the speed of the particle. mv qB q v = BR m 3(1.6 x 10 −19 C ) v= (0.25T )(0.475m) − 27 12(1.67 x 10 Kg ) R= v = 2.84 x106 m / s Example from Y+F In an experiment with cosmic rays, a vertical beam of particles that have a charge magnitude of 3e and a mass 12 times the proton mass enters a uniform horizontal Magnetic field of 0.250T and is bent in a semicircle of 95cm. Find the speed of the particle. mv R= qB q v = BR m m 3(1.6 x 10 −19 C ) v= (0.25T )(0.475m) − 27 12(1.67 x 10 Kg ) v = 2.84 x106 m / s Motion in a constant magnetic field If the velocity of the charge particle is perpendicular to the B field the motion is circle with radius R=mv/qB. If the velocity is not perpendicular, the motion is a helix. In this case we break up the velocity into components perpendicular v⊥ and parallel to the field v//. r r r F = qv × B = 0 Therefore have uniform motion in B direction. Path looks like extended spring, v// constant along the B direction and v⊥ moves in a circle. R (of helix) = mv⊥ /qB Circumference = 27 km = 2 π R LHC@Geneva, Switzerland R= mv p = qB qB The KEKB Collider (Tsukuba, Japan) 8 x 3.5 GeV beam energies Belle detector SCC RF(HER) World record: L = 2.1 x 1034/cm2/sec ARES(LER) Ares RF cavity e+ source 3.1 km around Applications Electric/Magnetic Velocity Selector A charged particle enters a region with perpendicular electric and magnetic fields. The electric and magnetic forces will cancel if the velocity is just right. Particles with this velocity will go through undeflected. Others will be deflected. r r r r r FE = qE = − Fm = −qv × B r r r qE = qv B1 r r E v = r B Choosing a particular E/B ratio will select the desired velocity. Example, E=10,000V/m and B=0.001T selects v=10,000,000 m/s Mass Spectrometer (Chemistry) Measure m to identify substances: 1.) Ionize atoms by hitting them with accelerated electrons. 2.) Acelerate ions through known potential V. q=e if singly ionized. U=qV 3.) Velocity select. 3.) Pass ions through known B field and measure R. R= mv qB 2 2 qB q B v= ; v2 = 2 2 mR m R 1 2 mv = qV 2 qB 2 R 2 m= 2V Mass Spectrometer (Chemistry) Applications: Think CSI Paleoceanography: Determine relative abundances of isotopes (they decay at different rates geological age) Space exploration: Determine what’s on the moon, Mars, etc. Detect chemical and biol. weapons (nerve gas, anthrax, etc.).