Classical Physics II PHY132 Lecture 13 Magnetism II: Magnetic torque Lecture 13 1
Magnetic Force MAGNETISM is yet another force that has been known since a very long time. Its name stems from the mineral magnetite, which is naturally magnetic. We now know that magnetism is the action-at-distance force felt between currents, i.e. ie moving charges magnetic materials contain currents, i.e. rapidly spinning and orbiting electrons, which align to give a macroscopic MAGNETIC FIELD because the magnetic forces are between (moving) charges, there must be a deep relationship with electricity ecause motion is dependent on the reference frame of the observer, we expect connections with relativity it as well Again, we use the FIELD concept: Currents (sources of ) )generate a Magnetic field everywhere in space, and interacts with the test -currents Lecture 13 2
Magnetic Force and Sources Note, that currents are by nature not point-like, but we will start with looking at the force between two very short pieces of wire carrying currents I 1 and I 2 This is necessarily an idealization! A current can only flow in a LOOP! y integration over the loops, we can calculate the total force between the currents and compare with experiment Currents in a wire have both a MAGNITUDE and a DIRECTION (// wire), i.e. currents are VECTORS by nature; unlike charges! Thus, the Magnetic Field (r) ( ) generated by a current I must depend on TWO vectors: position vector r and the current vector I VECTOR PRODUCT Note that experiment will ultimately decide if we are on the right track! Lecture 13 3
Magnetic Field from a Wire Current efore we skip to the EFFECTS of a magnetic field (r), we shortly describe the GENERATION of by a current; we ll return to this in great detail later... the simplest (?) case is the generation of a field by a infinitely long straight wire carrying a dl constant current I d d 0 Idl r iot-savart Law: I d 4 r 2 right-hand rule! gives the contribution d to the field due to the current I in a (infinitesimally) short piece of the wire dl in point r away from dl. d is to both dl and r! Integration over all the wire gives: r where is the unit vector to the wire: forms CIRCLES around the wire 0 Ir 2 r Lecture 13 4
Force on a Moving Charge We ve seen that t the magnetic field is generated by currents (i.e. vectors), and likewise interacts with currents and moving charges (i.e. vectors). Thus we expect the magnetic FORCE vector F exerted on a moving charge to depend on TWO vectors: and qv VECTOR PRODUCT: F = qv (Lorentz Force) Force F is perpendicular to both and v! In a wire carrying a current I = dq/dt, charge dq in the wire travels a short distance dl in time dt: dq v dq = dq dl/dt = Id dl Thus, the contribution to the total force from a short piece dl of a current-carrying wire is: df = dqv dq = Idl (Lorentz Force) for a straight wire of length l in a uniform field: F = Il Lecture 13 5
Units of Note that the magnetic FORCE is PERPENDICULAR to both and qv (or Idl): right-hand hand rule! i.e. the Magnetic field is NOT the direction of the magnetic force!! Acceleration: a // = 0; only a perp 0 Inspecting the formula F = qv, we see that t there is NO PROPORTIONALITY constant! Strength of is defined as Force-per-unit-current-meter or equivalently as Force-per-unit-charge-speed remember: E was defined as Force-per-unit-test-charge indeed the UNIT of is defined in terms of the other SI variables: [] = N/(Cm/s) = N/(A m) T(esla) older unit: G(auss) = 10 4 T Lecture 13 7
Example: proton in a -Field A proton (m=1.67 10 27 kg, q=e=1.6 10 19 ) enters a uniform magnetic field v: v If v then the force F,, which is perpendicular p to both, causes the particle to turn Acceleration a is ALWAYS to the path: a is a CENTRIPETAL acceleration, and the particle will go through a circular/helical path with constant speed v e.g. v = 3 10 7 m/s, = 1.5 T; the trajectory: q q F F v a m v m 2 v qv sinv, qv a v, R m m 27 7 mv 1.6710 310 R 21cm 19 q 1.610 1.5 a The ANGULAR VELOCITY (cyclotron frequency) of the proton (while 7 in the field) equals: v q 310 8 2 f 1.410 rad/s R m 0.21 F F R F Lecture 13 8 F F
A ubble Chamber Picture Practical tool to discover and measure properties of elementary particles: use of magnetic forces on a fastmoving particles, as protons or pions: π p K 0 Λ 0 K 0 π π, Λ 0 pπ π π K 0 Λ 0 π + p 16 cm π 16 cm Lecture 13 10
Electron in E and Fields Consider an electron (q = e, m) ) of velocity v entering a region of mutually perpendicular E and fields: we can arrange the strength of the fields such that the NET force on the electron is zero: FqEqv0 v Q ve V F F F electron source q = e v +Q F E F E F E E E when we put a slit behind the system, we get a velocity filter if we put an accelerator U = qv = ½ mv 2 in front we measure the ratio q/m of the electron: 2 q v 1 E m 2V with velocity filter 2V Lecture 13 11 2
Force on a Current Loop Consider a rigid id rectangular wire loop, carrying a current I, in an uniform magnetic field. Let = i, and sides da and bc parallel l to the y-axis (into the paper): the loop (i.e. the normal A to the plane) makes an angle θ with ih F Net Force = 0! Net Torque 0: Fab Fcd 0 Fbc Idl F 1 1 bc Ibc τ abf cd F abf bc da bc 2 2 abfbcsin ab bc I sin AIsin μ IA τ μ F cd c F bc with ih μ = IA the magnetic i moment of the loop This can easily be generalized to any arbitrary (flat) loop b d F da θ I a A x F ab right-hand rule: curl fingers around the loop in I direction, thumb is direction of A Lecture 13 12
Example Consider a flat round wire loop of n = 20 turns of wire, radius R = 20 cm, carrying a current of I = 0.50 A. A uniform 1.0T-Field is parallel to the plane of the loop: = i I μ = nia=nia j I b a Calculate the torque on the loop: τ μnia niai j nia k 2 20 0.50 0.20 1.0 k 1.26 Nm (out of paper) Thus, the a-side goes up, and the b-side down: the loop tends to rotate around the z-axis to a position of minimum potential energy, i.e. alignment of its magnetic moment μ and Suspend the loop free to rotate around the z-axis, keep it in equilibrium with small torque springs a spool ampere-meter Lecture 13 14
Potential ti Energy: Potential Energy of a Loop F da U d τ dθ I /2 /2 τ opposes increase in θ choosing U=0 for θ=90 cos /2 AIsin d AI AI cos μ F cd c b F bc θ a x F ab A μ =IA zero (with this choice of zero-point for U) for μ and perpendicular minimum occurs when μ and are parallel: U = μ maximum mum occurs when μ and are anti-parallel: U = +μμ Lecture 13 15
Example The plane of an A = 4.0 cm 16.0 cm rectangular wire loop is parallel to a = 0.30 T i magnetic field. The loop carries an I = 50Acurrent 5.0 current, such that the magnetic moment vector μ = IA is along the y-axis: Calculate the loop s magnetic moment and the torque acting on the loop: 4 2 2 μ IA IAj5.06410 j3.210 jam τ μ ji k k 32 2 3.2 10 030 0.30 96 9.6 10 3 Nm(intopaper) Calculate the maximum possible torque on the loop for the same current, magnetic field, and total wire length: The torque is maximum when the area of the loop is maximum: for a loop with circumference of 2πR = 40 cm A = πr 2 = 400/π = 127 cm 2 τ max = IA = 6.4 10 2 Nm 2 Calculate the magnetic potential energy if the normal to the loop makes a 30 angle with the field 2 3 U μ3.210 0.30sin 304.810 Nm Lecture 13 16