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Review on Magnetism Chapter 28 Magnetism 1 Magnetism • Refrigerators are attracted to magnets! Magnetism 2 Where is Magnetism Used?? • Motors • Navigation – Compass • Magnetic Tapes – Music, Data • Television – Beam deflection Coil • Magnetic Resonance Imaging (MRI) • High Energy Physics Research Magnetism 3 FE (28 – 8) FE qE Cathode FB qv B Anode FB Magnetism 4 Consider a Permanent Magnet B N S The magnetic Field B goes from North to South. Magnetism 5 Units F Bqv Sin(θ ) Units : F N N B qv Cm / s Amp m Magnetism 1 tesla 1 T 1 N/(A - m) 6 Typical Representation Magnetism 7 A Look at the Physics B q v q B There is NO force on a charge placed into a magnetic field if the charge is NOT moving. There is no force if the charge moves parallel to the field. • If the charge is moving, there is a force on the charge, perpendicular to both v and B. F=qvxB Magnetism 8 The Lorentz Force This can be summarized as: F qv B F or: F qvBsin v B mq is the angle between B and V Magnetism 9 Nicer Picture Magnetism 10 The Wire in More Detail Assume all electrons are moving with the same velocity vd. L L q it i vd F qvd B i L vd B iLB vd vector : F iL B B out of plane of the paper Magnetism Vector L in the direction of the motion of POSITIVE charge (i). 11 (28 – 12) Magnetic force on a straight wire in a uniform magnetic field. If we assume the more general case for which the magnetic field B froms and angle with the wire the magnetic force equation can be written in vector form as: FB iL B FB iL B B i dF . dFB = idL B FB i dL B Magnetism Here L is a vector whose magnitude is equal to the wire length L and has a direction that coincides with that of the current. The magnetic force magnitude FB iLB sin dL Magnetic force on a wire of arbitrary shape placed in a non - uniform magnetic field. In this case we divide the wire into elements of length dL which can be considered as straight. The magnetic force on each element is: dFB = idL B The net magnetic force on the wire is given by the integral: FB i dL B 12 Current Loop What is force on the ends?? Loop will tend to rotate due to the torque the field applies to the loop. Magnetism 13 (28 – 13) Side view Top view C C net iAB sin Fnet 0 Magnetism 14 Dipole Moment Definition Define the magnetic dipole moment of the coil m as: m=NiA =m x B Magnetism We can convert this to a vector with A as defined as being normal to the area as in the previous slide. 15 mB U mB U m B U m B Magnetic dipole moment : The torque of a coil that has N loops exerted by a uniform magnetic field B and carrries a current i is given by the equation: NiAB We define a new vector m associated with the coil which is known as the magnetic dipole moment of the coil. Magnetism 16 (28 – 14) R L The Hall effect In 1879 Edwin Hall carried out an experiment in which R he was able to determine that conduction in metals is due to the motion of negative charges (electrons). He was also L able to determine the concentration n of the electrons. He used a strip of copper of width d and thickness . He passed a current i along the length of the strip and applied a magnetic field B perpendicular to the strip as shown in the figure. In the presence of B the electrons experience a magnetic force FB that L R pushes them to the right (labeled "R") side of the strip. This accumulates negative charge on the R-sid e and leaves the left side (labeled "L") of the strip positively charged. As a result of the accumulated charge, an electric field E is generated as shown in the figure so that the electric force balances the magnetic force on the moving charges. FE FB eE evd B E vd B (eqs.1). From chapter 26 we have: J nevd Magnetism vd J i i ne Ane dne (eqs.2) 17 (28 – 15) Motion of a charged particle in a magnetic Field Magnetism 18 Trajectory of Charged Particles in a Magnetic Field (B field points into plane of paper.) + +B + v+ + + + + + + + + F + + + + F + + + + + + + + + + + + + + + + + + + B + + + + + Magnetism v 19 Trajectory of Charged Particles in a Magnetic Field (B field points into plane of paper.) + +B + v+ + + + + + + + + F + + + + F + + + + + + + + + + + + + + + + + + + B + + + + + Magnetism v Magnetic Force is a centripetal force 20 Review of Rotational Motion = s / r s = r ds/dt = d/dt r v = r s r = angle, = angular speed, = angular acceleration at ar at = r tangential acceleration ar = v2 / r radial acceleration The radial acceleration changes the direction of motion, while the tangential acceleration changes the speed. Uniform Circular Motion ar = constant v and ar constant but direction changes v Magnetism ar = v2/r = 2 r KE = ½ mv2 = ½ mw2r2 F = mar = mv2/r = m2r 21 Magnetism 22 Radius of a Charged Particle Orbit in a Magnetic Field +B + + v+ + + + + + + + + + r + + + + + + F + Magnetism Centripetal Force = Magnetic Force mv 2 qvB r mv r qB 23 Cyclotron Frequency +B + v+ + + + + + + + + + r + + + + + + + F + Magnetism The time taken to complete one orbit is: 2r T v 2 mv v qB 1 qB f T 2 m qB c 2f m 24 Mass Spectrometer Smaller Mass Magnetism 25 Magnetism 26 An Example A beam of electrons whose kinetic energy is K emerges from a thin-foil “window” at the end of an accelerator tube. There is a metal plate a distance d from this window and perpendicular to the direction of the emerging beam. Show that we can prevent the beam from hitting the plate if we apply a uniform magnetic field B such that 2mK B 2 2 ed Magnetism 27 Problem Continued r From Before mv r qB 1 2 2K K mv so v 2 m m 2K 2mK r d 2 2 eB m e B Solve for B : 2mK B e2d 2 Magnetism 28 #14 Chapter 28 A metal strip 6.50 cm long, 0.850 cm wide, and 0.760 mm thick moves with constant velocity through a uniform magnetic field B= 1.20mTdirected perpendicular to the strip, as shown in the Figure. A potential difference of 3.90 ηV is measured between points x and y across the strip. Calculate the speed v. FIGURE 2837 Magnetism Problem 14. 29 21. (a) Find the frequency of revolution of an electron with an energy of 100 eV in a uniform magnetic field of magnitude 35.0 µT . (b) Calculate the radius of the path of this electron if its velocity is perpendicular to the magnetic field. Magnetism 30 39. A 13.0 g wire of length L = 62.0 cm is suspended by a pair of flexible leads in a uniform magnetic field of magnitude 0.440 T. What are the (a) magnitude and (b) direction (left or right) of the current required to remove the tension in the supporting leads? Magnetism 31