Magnetic field and force So far we have studied two forces: gravity and electricity Magnetism is a new force, but also related to electric charges. Gravity is created by mass and gravity acts on masses. v kQ Electric fields are created by electric charges E = 2 rˆ v r v And Electric fields exert forces on charges. FE = qE Magnetic field and force There is a different kind of field, called a magnetic field, or B-field v B B-fields are created by moving charges (currents). B-fields exert a force on moving charges. This is very different from our previously studied forces! 1 Natural magnetism Natural magnetism 2 What is the origin of this force? Magnetic charge? Monopoles? Scientists have postulated that there might be some other kind of object that has a “magnetic charge” = Magnetic Monopole. No magnetic monopoles have ever been found. Thus, we do not consider them in classes. 3 Magnetism and electric charge Oersted’s experiment: A compass is placed directly over a wire (here viewed from above). A passing current deflects the needle. Force on a moving charge We will see later exactly how B-fields are made by moving charges. Now we will study how B-fields exert forces on moving charges. v v v F = qv × B The magnetic force exerted on a charge q moving with velocity v in a magnetic field B. This equation actually defines the magnetic B-field. 4 Visualizing a M. Field: Field lines http://www.youtube.com/watch?v=wuA-dkKvrd0 Force on a moving charge v v v F = qv × B Vector cross product. v v v | F |= q | v || B | sinθ What about the resulting force direction from B the cross product? θ v 5 Vector cross productv v– Right Hand Rule u × v • Fingers in direction of first vector. • Bend them into direction of second vector. •Thumb points in cross product direction. •If your hand does not bend that way, flip it around! – never use the left hand ;-) New vector direction is always perpendicular to the original vectors! Force on a moving charge 6 Consequences: v v v F = qv × B v v v | F |= q | v || B | sinθ • If particle is not moving (v=0) then no force. • If velocity and B-field are parallel, then no force. • If velocity and B-field are perpendicular, then maximum force. • If q is negative, then force is in opposite direction. Drawing conventions Drawing vector directions on a 2 dimensional piece of paper? Magnetic Field Vector to the right. Magnetic Field Vector out of the page. Arrow head pointing at you. Magnetic Field Vector into the page. X Arrow tail pointing at you. 7 Clicker Question A negative particle and a positive particle are moving with certain velocities in a constant, uniform magnetic field, as shown. The direction of the B-field is to the right. The (+) particle is moving directly left; the (–) particle is moving directly up. The force on the positive particle due B to the B-field is (in = into page, out = out of page). A: in B: out C: zero D: right E: left Answer: The (+) particle is moving anti-parallel to the B-field. The angle θ is 180 and the force is FB=qvB sinθ = 0. Clicker Question A negative particle and a positive particle are moving with certain velocities in a constant, uniform magnetic field, as shown. The direction of the B-field is to the right. The (+) particle is moving directly left; the (–) particle is moving directly up. The force on the negative particle B due to the B-field is A: in B: out C: zero D: right E: left Answer: The (–) particle is moving at right angles to the field. By the right-hand rule, the direction "v cross B" is into the page, but the particle has a negative charge q, so the force is out of the page. 8 Clicker Question A positive particle is released from rest in a region of space where there is constant, uniform, electric field and a constant, uniform magnetic field. The A B C electric field points up and the E magnetic. field points out of the B(out) page in the diagram below. Which path will the positive particle follow? (All paths shown are in plane of the page.) D: it will not move Answer: The (+) particle will feel a force FE = qE due to the E- field along the direction of the E-field. As it starts moving along the E-field direction, it will acquire a velocity, and it will start to feel a force FB=qvB, due to the B-field. The direction of the force is to the right, by the right-hand-rule. Clicker Question A negative particle and a positive particle are moving with certain velocities in a constant, uniform magnetic field, as shown. The direction of the B-field is to the right. The (+) particle is moving directly left; the (–) particle is moving directly up. The force on the negative particle B due to the B-field is A: in B: out C: zero D: right E: left Answer: The (–) particle is moving at right angles to the field. By the right-hand rule, the direction "v cross B" is into the page, but the particle has a negative charge q, so the force is out of the page. 9 Magnetic field units v v v Units for Magnetic Field F = qv × B B-field = [B] = [Newtons] / [Coulomb x meters/second] = [Tesla] How big is a 1 Tesla Magnetic Field? Interstellar Space 10-10 Tesla Human Being 10-10 Tesla Earth’s Surface 5x10-5 Tesla Sun’s Surface 10-2 Tesla Small Bar Magnet 10-2 Tesla Experiment Magnet 1 Tesla Maximum Steady Magnet 30 Tesla Maximum in Explosive Magnet 1000 Tesla Surface of Neutron Star 108 Tesla 10 Another Unit System 1 Gauss = 10-4 Tesla Thus, the Earth’s magnetic field is ~ 0.5 Gauss. This unit system is often used when talking about small magnetic fields, but it is not the SI unit system! Earth’s magnetic field 11 Clicker Question Here is an event display from a high energy experiment. There is a 1 Tesla uniform magnetic field coming out of the page. What is the sign of the electric charge? v v v A)Positive B)Negative F = qv × B Videos Feymann's "why" and "Fields" 12 Motion of charged particles in a M. Field v v v Because the force is always F qv B perpendicular to the velocity (direction = × of motion), the Magnetic force can do no work on q. v v v v since W = F ⋅∆r = 0 FB ⊥ ∆r B-field cannot change the Kinetic Energy of a moving particle, but can change its direction of motion. Wnet = ∆KE = 0 Charged particle in a perpendicular field Particle moving in a plane with a B- v field uniformly out of the plane. B Results in circular motion! v F No change in KE, but constant change in velocity direction. B F 13 Charged particle in a perpendicular field v v v v | F |= m | a |=| qv × B | v v v | F |= q | v || B | Since velocity and B are always perpendicular. v v v mv2 | F |= q | v || B |= Since circular motion. R v mv Radius of circular motion R R = depends on m, v, q, B. qB B Charged particle in a perpendicular field One can then solve for the frequency of v revolution (“cyclotron frequency”). R f =# revolutions/second B qRB distance 2πR v = = = m time T Period 1 qB f = = The cyclotron frequency is T 2πm independent of R. 14 Helical motion What if in addition to the v mv circular motion in this plane, R R = there is a non-zero velocity qB out of the page? B This extra velocity contributes no additional force since it is parallel to B. Aurora borealis (“Northern lights”) 15 Clicker Question A (+) charged particle with an initial speed vo is moving in a plane perpendicular to a uniform magnetic field (B into the page). There is a tenuous gas throughout the region which causes viscous drag and slows the particle over time. The path of the particle is B A: a spiral inward B: a spiral outward C: something else Applications: Mass spectrometer 16 Applications: velocity selector Electric force: FE = qE Magnetic force: FB = qvB Total force: ∑ F =qvB − qE For the particle to pass and not be deflected: E qvB − qE = 0 → v = B Thomson’s e/m experiment 1 2eV From energy conservation: mv 2 = eV → v = 2 m E We have seen that to pass v = through we need: B e E 2 = Combining the two we obtain: m 2VB2 17 Force on current carrying wires v v Since B-fields exert forces on moving v charges, it is natural that B-fields exert F = qv × B forces on current carrying wires. How do we quantify this force in terms of current I, instead of q and v? Force on current carrying wires v v v Consider the wire shown below. F(one charge) = qv × B Now add a Magnetic Field. # charges in this Force segment of wire X B N = n x A x L Velocity v + Area A #/Volume Volume v v v Length vector L F(total) = (nALq)v × B 18 Force on current carrying wires v v v F(total) = (nALq)v × B How to relate this to current? I J = = nqv I = nAqv A d d v v v F(total) IL B * Notice that L points along v, which = × is the direction of the current.