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CHAPTER 27 Magnetic Field and Magnetic Forces PRODUCING A MAGNETIC FIELD Permanent Magnet Current MAGNETIC FIELD LINES, BAR MAGNET EXAMPLE The compass can be used to trace the field lines The lines outside the magnet point from the North pole to the South pole MAGNETIC FIELD LINES, BAR MAGNET MAGNETIC FIELD LINES OF CURRENT GEOGRAPHIC NORTH VS MAGNETIC NORTH MAGNETIC FIELD B Unit: T (Tesla) 1T=1Ns/(Cm) Magnetic field of earth: 10-4T (Reverses every 250 000 years) Poles moves 30 miles per year Non SI unit: Gauss (G) 1 G = 10-4T FORCE ON A CHARGE FB = qv ´ B (Vector cross product) Þ FB = qvBsinf (Magnitude) f is the angle between v and B. Special cases : v = 0 Þ FB = 0 f = 0 or 180 Þ FB = 0 f = 90 Þ FB = qvB DIRECTION OF MAGNETIC FORCE FB = qv ´ B q>0 q<0 RIGHT HAND RULE EXAMPLES OF DIRECTION OF MAGNETIC FORCE EXAMPLES OF DIRECTION OF MAGNETIC FORCE MAGNETIC FORCE ON A PROTON Find the magnetic force on a proton (q = 1.6 ´ 10 -19 C) moving at 3 ´ 10 5 m /s through a uniform magnetic field of 2T. Magnitude : F = qvBsin f = (1.6 ´ 10 -19 C)(3 ´ 10 5 m /s)(2T)sin 30 = 4.8 ´ 10 -14 N Right hand rule tells us the force is point in the negative y - axis. VECTOR CROSS PRODUCT ìïu = ux iˆ + uy ˆj + uz kˆ í ïîv = v x iˆ + v y ˆj + v z kˆ u ´ v º (uyv z - uzv y )iˆ + (uzv x - ux v z ) ˆj + (uxv y - uyv x ) kˆ iˆ = ux vx ˆj uy vy kˆ uz (determinant) vz Useful facts : (1) u ´ v = - v ´ u (2) u ´ v = uv sin q where u = u , v = v q is the angle between u and v . CROSS PRODUCT EXAMPLE ìu = - iˆ + 2 ˆj + 3kˆ í ˆ + 5 ˆj - 6kˆ v = 4 i î u ´ v = (uyv z - uzv y )iˆ + (uzv x - ux v z ) ˆj + (uxv y - uyv x ) kˆ iˆ ˆj kˆ = -1 2 3 = -27iˆ + 6 ˆj - 13kˆ 4 5 -6 VECTOR CROSS PRODUCT iˆ ´ ĵ = k̂ iˆ ´ iˆ = 0 k̂ If C = A ´ B, then we have: C^A iˆ ĵ and C^B EXAMPLE q = -1C,v = (2 ĵ)m / s, B = (3iˆ - 2 ĵ - 9 k̂)T What is F ? k̂ F = qv ´ B = (-1)(2 ĵ) ´ (3iˆ - 2 ĵ - 9 k̂)N = (-6 ĵ ´ iˆ + 4 ĵ ´ ĵ + 18 ĵ ´ k̂)N = (6 k̂ + 0 + 18iˆ )N = (18iˆ + 6 k̂)N iˆ ĵ ANOTHER METHOD F = qv ´ B = (-1)(2 ĵ) ´ (3iˆ - 2 ĵ - 9 k̂)N iˆ ĵ k̂ = (-1) 0 2 0 N 3 -2 -9 = (18iˆ + 6 k̂)N IN AND OUT OF THE PLANE Symbols for B field going in/out of paper Going into the paper Going out of the paper EXAMPLE B = 1.2mT pointing out. v Proton : v = 3.2 ´ 10 7 m /s , m = 1.67 ´ 10 -27 kg, q = 1.6 ´ 10 -19 C What is the force and acceleration. What is the angle f? f = 90 Þ sin f = 1 F = qvBsin f = 6.1 ´ 10 -15 N Points to the RIGHT a = F / m = 3.7 ´ 1012 m / s 2 B EXAMPLE v What if ϕ is not 90o ? f = 30 Þ sin f = 1 / 2 F = qvBsin f = 3 ´ 10 -15 N Points into the screen a = F / m = 1.8 ´ 10 m / s 12 2 30o B CIRCULAR MOTION IN A MAGNETIC FIELD What happens when a positively charged particle enters with velocity v into a region with B field pointing down (into the screen). FB = qvB mv 2 Fcentripetal = r mv 2 Þ qvB = r mv Þr= qB BENDING OF AN ELECTRON BEAM MEASURING CHARGE TO MASS RATIO Q/M Measuring r gives you q/m. mv r= qB q v Þ = m Br What if the charge is negative? A BETTER WAY TO GET Q/M The velocity v is often difficult to measure, but we can figure out v using the potential difference V. From the first equation : qBr v= m Put into the second : 1 qBr 2 m( ) = qV 2 m q B 2r 2 Þ =V m 2 q 2V Þ = 2 2 m Br ì mv 2 ï r = qvB í1 ï mv 2 = qV î2 FIND THE PARTICLE VELOCITY Mass : 12m p Charge : q = 3e Field : 0.25T (m p = 1.67 ´ 10 -27 kg, e = 1.6 ´ 10 -19 C) Charge is negative. r = 95cm /2 = 0.475m qBr 3eBr v= = = 2.84 ´ 10 6 m /s m 12m p CROSSED FIELDS q vB = q E when there is not deflection Þv= E/B FORCE ON CURRENT F = IL ´ B (V ector) Þ F = ILBsinf (Magnitude) Does it matter whether the charge carriers are positive or negative? EXAMPLES OF MAGNETIC FORCE ON A WIRE F = IL ´ B DERIVATION: MAGNETIC FORCE ON A WIRE If there are N number of charges in a wire of length L, each with charge q moving at v d . The force on the wire is : F = Nqv d ´ B But J = nqv d , n : number of charge per unit volume N Þ J = ( )qv d AL Þ Nqv d = ALJ = IL, L points along the current. Put together, we have : F = IL ´ B MAGNETIC DIPOLE MOMENT A loop of current carries magnetic dipole moment (or magnetic moment): m = IA SI unit: Am2 The direction of the area vector is determined by the right hand rule. DIPOLE IS A SOURCE OF B FIELD The B field generated by a magnetic dipole is roughly the same as that of a bar magnet. μ EXAMPLE Find the magnetic dipole of a loop of wire of radius 10cm carrying a current of 3A. m = IA = 3p r 2 = 0.094 Am2 What if the wire has N=30 loops? m = NIA = 30(3p r 2 ) = 2.8Am2 TORQUE ON A DIPOLE t =m´B Magnitude : t = m ´ B = mBsinf μ and B tends to align m EXAMPLE: TORQUE ON A DIPOLE POTENTIAL ENERGY OF A DIPOLE U = -m × B = -mBcosf μ and B tends to align in order to attain the lowest energy. m