Physics 202 1436-1437 H Instructor: Dr. Alaa Mahmoud E-mail: alaa_y_emam@hotmail.com
Chapter 28 magnetic Field
Magnetic fingerprinting allows fingerprints to be seen on surfaces that otherwise would not allow prints to be lifted. The powder spread on the surface is coated with an organic material that adheres to the greasy residue in a fingerprint. A magnetic brush removes the excess powder and makes the fingerprint visible. (James King-Holmes/Photo Researchers, Inc.)
What Produces a Magnetic Field? Electric field E is produced by an electric charge Magnetic field is produced by? (there is NO magnetic charge) individual (imaginary) magnetic charges is called magnetic monopoles How then are magnetic fields produced? There are two ways: 1. Moving electrically charged particles, such as a current in a wire, to make an electromagnet 2. The elementary particles such as electrons have an intrinsic magnetic field around them (basic characteristic of each particle) For some materials, magnetic fields of the electrons add together to give a net magnetic field around the material à permanent magnet Other materials, the magnetic fields of the electrons cancel out, giving no net magnetic field surrounding the material. Such cancellation is the reason you do not have a permanent field around your body
The Definition of We define in in terms of the magnetic force F exerted on a moving electrically charged test particle by firing a charged particle (of velocity v ) through the point at which is to be defined, and determining the force F that acts on the particle at that point F : qv : : ; The magnitude of F : F q v sin f,, where! is the angle between v & F never has a component parallel to v à cannot change the particle s speed v F changes only the direction of v à the particle accelerate by F F = 0 if v & are either parallel (!= 0 ) or antiparallel (! =180 ), F is maximum when v & are perpendicular to each other Finding the Magnetic Force on a Particle The direction of F is always normal to both velocity and magnetic field, determined by right-hand rule v v v v F F (a) (b) (c) (d) (e)
The SI unit for is tesla (T) 1 tesla 1 T 1 coulomb per second is an ampere, we have 1 tesla 10 4 gauss. newton (coulomb)(meter/second). 1 newton (coulomb/second)(meter) 1 N A m. CHECKPOINT 1 The figure shows three y y y situations in which a charged particle with velocity v : v travels through + x x a uniform magnetic field : z z z. In each situation, v v what is the direction of the magnetic force F : (a) (b) (c) on the particle? x a) +z; b) -x; c) F =0
Magnetic fields lines: (1) the direction of the tangent to a magnetic field line at any point gives the direction of (2) the spacing of the lines represents the magnitude of (stronger where the lines are closer together) Magnetic field lines of bar magnet: The lines all pass through the magnet, and they all form closed loops The external magnetic effects of a bar magnet are strongest near its ends, where the field lines are most closely spaced lines goes from the north pole N of the the south pole S ecause a magnet has two poles, it is said to be a magnetic dipole N S N S N S The field lines run from the north pole to the south pole. ) Opposite magnetic poles attract each other, and like magnetic poles repel each other.
Sample Problem Magnetic force on a moving charged particle A uniform magnetic field, with magnitude 1.2 mt, is directed vertically upward throughout the volume of a laboratory chamber. A proton with kinetic energy 5.3 MeV enters the chamber, moving horizontally from south to north. What magnetic deflecting force acts on the proton as it enters the chamber? The proton mass is 1.67 10 27 kg. (Neglect Earth s magnetic field.) ecause the proton is charged and moving through a magnetic field, a magnetic force F : can act on it. ecause the initial direction of the proton s velocity is not along a magnetic field line, is not simply zero. F : Magnitude: To find the magnitude of F :, we can use Eq. 28-3 (F q v sin ) provided we first find the proton s speed v. We can find v from the given kinetic energy because K 1 2 mv2.solving for v,we obtain v A 2K m A 3.2 10 7 m/s. Equation 28-3 then yields F q v sin (1.60 10 19 C)(3.2 10 7 m/s) (1.2 10 3 T)(sin 90 ) 6.1 10 15 N. : KEY IDEAS (2)(5.3 MeV)(1.60 10 13 J/MeV) 1.67 10 27 kg (Answer) This may seem like a small force, but it acts on a particle of small mass, producing a large acceleration; namely, a F m 6.1 10 15 N 1.67 10 27 kg 3.7 1012 m/s 2. Direction: To find the direction of, we use the fact that has the direction of the cross product qv : :. ecause the charge q is positive, F : must have the same direction as v : :, which can be determined with the right-hand rule for cross products (as in Fig. 28-2d). We know that v : is directed horizontally from south to north and : is directed vertically up. The right-hand rule shows us that the deflecting force F : must be directed horizontally from west to east, as Fig. 28-6 shows. (The array of dots in the figure represents a magnetic field directed out of the plane of the figure.an array of Xs would have represented a magnetic field directed into that plane.) If the charge of the particle were negative, the magnetic deflecting force would be directed in the opposite direction that is, horizontally from east to west. This is predicted automatically by Eq. 28-2 if we substitute a negative value for q. W N v + S F F : Path of proton Fig. 28-6 An overhead view of a proton moving from south to : north with velocity v in a chamber. A magnetic field is directed vertically upward in the chamber, as represented by the array of dots (which resemble the tips of arrows). The proton is deflected toward the east. E F :
A Circulating Charged Particle v If a charged particle q of mass m and speed v enters a uniform à the particle will move in a circular path of radius r given by r mv q (radius). F The period T (the time for one full revolution) is equal to the circumference divided by the speed: T 2 r v 2 v mv q 2 m q (period). (28-17) The frequency f (the number of revolutions per unit time) is f 1 T q 2 m (frequency). (28-18) The angular frequency v of the motion is then 2 f q m (angular frequency). (28-19)
Magnetic force on a wire carrying current 8_735-763v2.qxd 27-11-2009 16:19 Page 751 Magnetic Force on a Current-Carrying Wire Consider a length L of the wire All conduction electrons will drift past plane xx in a time t L/vd. Thus, a charge q given by q it i L v d F qv d sin il v d v d sin 90 F il. The magnetic force due to a current passing through a wire of length L is L i F F : il : : (force on a current). v d CHECKPOINT 4 The figure shows a current i through a wire in a uniform magnetic field,as well as the magnetic force F : acting on the wire.the field is oriented so that the force is maximum. In what direction is the field? y : x x i x z F Answer: -y Sample Problem
Sample Problem Magnetic force on a wire carrying current A straight, horizontal length of copper wire has a current i 28 A through it. What are the magnitude and direction of the minimum magnetic field : needed to suspend the wire that is, to balance the gravitational force on it? The linear density (mass per unit length) of the wire is 46.6 g/m. KEY IDEAS (1) ecause the wire carries a current, a magnetic force can act on the wire if we place it in a magnetic field :. To balance the downward gravitational force F : on the wire, we want F : g to be directed upward (Fig. 28-17). (2) The direction of F : is related to the directions of : and the wire s length vector L : by Eq. 28-26 (F : il : : ). Calculations: ecause is directed horizontally (and the current is taken to be positive), Eq. 28-26 and the right-hand rule for cross products tell us that : must be horizontal and rightward (in Fig. 28-17) to give the required upward F :. The magnitude of F : is F il sin f (Eq. 28-27). ecause we want to balance, we want F : L : where mg is the magnitude of F : il sin f mg, (28-29) F : g F : g and m is the mass of the wire. Fig. 28-17 of the page. A wire (shown in cross section) carrying current out We also want the minimal field magnitude for F : to balance F : g.thus,we need to maximize sin f in Eq. 28-29. To do so, we set f 90, thereby arranging for : to be perpendicular to the wire.we then have sin f 1, so Eq. 28-29 yields L F mg mg (m/l)g. il sin i (28-30) We write the result this way because we know m/l,the linear density of the wire. Substituting known data then gives us (46.6 10 3 kg/m)(9.8 m/s 2 ) 28 A 1.6 10 2 T. (Answer) This is about 160 times the strength of Earth s magnetic field.
Torque on a Current Loop F i The forces behind electric motors work are the magnetic forces a simple motor, consisting of a single current-carrying loop immersed in The two magnetic forces F & -F produce a torque on the loop à tending to rotate it about its central axis N i F S The torque " tends to rotate the loop so as to align its normal vector with the direction of If we replace the single loop of current with a coil of N loops, or turns The total torque on the coil then has magnitude for a circular coil, with radius r, we have t (NiA) sin u, t (Nipr 2 ) sin u. In a motor, the current in the coil is reversed, so that a torque continues to rotate the coil
The Magnetic Dipole Moment If a current i passes through a coil (wire) of turns N, area A à a current-carrying coil is said to be a magnetic dipole The magnetic dipole moment #: The direction of #: normal to the plane of the coil and is given by the same right-hand rule The magnitude of #: is given by m NiA (magnetic moment), N :is the number of turns in the coil, i :is the current through the coil A is the area enclosed by each turn of the coil The SI unit of magnetic dipole is A.m 2
If this coil is placed in a uniform external magnetic field à a torque will be experienced as t (NiA) sin u, A magnetic dipole in an external magnetic field has an potential energy U that depends on the dipole s orientation in the field If $ = 0, U = - % (the dipole moment is lined up with the magnetic field If $ = 180, U = + % (the dipole moment is directed opposite the field the unit of % is also joule per tesla (J/T) t m sin u, : : :, U( ) : :. The magnetic moment vector attempts to align with the magnetic field. i µ Highest energy µ Lowest energy i If an applied torque (due to an external agent ) rotates a magnetic dipole from an initial orientation $ i to another orientation $ f, à work W a is done on the dipole by the applied torque. If the dipole is stationary before and after the change in its orientation, then W a U f U i,
CHECKPOINT 5 The figure shows four orientations, at angle u, of a magnetic dipole moment in a magnetic field. Rank the orientations according to (a) the magnitude of the torque on the dipole and (b) the orientation energy of the dipole, greatest first. : 1 µ µ 2 θ θ θ θ 4 µ µ 3 (a) all tie; (b) 1 and 4 tie, then 2 and 3 tie
Sample Problem Rotating a magnetic dipole in a magnetic field Figure 28-21 shows a circular coil with 250 turns, an area A of 2.52 10 4 m 2,and a current of 100mA.The coil is at rest in a uniform magnetic field of magnitude 0.85 T, with its magnetic dipole moment : initially aligned with. (a) In Fig. 28-21, what is the direction of the current in the coil? Right-hand rule: Imagine cupping the coil with your right hand so that your right thumb is outstretched in the direction of :. The direction in which your fingers curl around the coil is the direction of the current in the coil. Thus, in the wires on the near side of the coil those we see in Fig. 28-21 the current is from top to bottom. (b) How much work would the torque applied by an external agent have to do on the coil to rotate it 90 from its ini- Fig. 28-21 A side view of a circular coil carrying a current and oriented so that its magnetic dipole moment is aligned with magnetic field :. : µ tial orientation, so that : is perpendicular to and the coil is again at rest? The work W a done by the applied torque would be equal to the change in the coil s orientation energy due to its change in orientation. Calculations: From Eq. 28-39 (W a U f U i ), we find W a U(90 ) U(0 ) cos 90 ( cos 0 ) 0. Substituting for m from Eq. 28-35 (m NiA), we find that W a (NiA) KEY IDEA (250)(100 10 6 A)(2.52 10 4 m 2 )(0.85 T) 5.355 10 6 J 5.4 J. : (Answer)
Problems: 1. An electron of a speed of 2 10 5 m/s perpendicularly enters a uniform magnetic field of 4 mt. Calculate (i) the magnetic force and (ii) the radius of its path. Solution (i) The magnitude of the magnetic force, since the velocity is normal to magnetic field, is F qv 1.6 10 19 2 10 5 4 10 3 1.28 10 16 N (ii) The radius of its circular path is 31 5 R mv 9.11 10 2 10 q 1.6 10 19 4 10 3 2.85 10 4 m
2. In a certain electric motor wires that carry a current of 8 A are perpendicular to a magnetic field of 50 mt. Calculate the magnetic force on each centimeter of these wires. Solution The magnitude of the magnetic force, since the magnetic field is normal to the wire, is F il 8 0.01 50 10 3 0.004N
3. A proton moving with a speed of 4 10 6 m/s through a magnetic field of 4 T experiences a magnetic force of 12.8 10-13 N. Find the angle between the proton s velocity and the magnetic field. Solution The magnitude of the magnetic force is defined as F qvsin Hence the angle is F sin 12.8 10 13 0.5 qv 1.6 10 19 4 10 6 4 30 0
4. A proton moves with a speed of 4 10 6 m/s normally to a magnetic field of 2.4 T. Calculate the acceleration of the proton due to the magnetic force. Solution The magnitude of the magnetic force is F qv 1.6 10 19 4 10 6 2.4 1.54 10 12 N Hence the magnitude of its acceleration (from Newton s second law) is F ma a F 1.54 10 12 9.2 10 14 m / s 2 m 1.67 10 27
5. An electron is in equilibrium under the influence of magnetic and electric forces. If the magnitude of the magnetic field is 1.2 T and its speed is 5 10 4 m/s, calculate the magnitude of the electric field. Solution The magnitude of the electric field is v E E v 5 10 4 1.2 6 10 4 V / m
6. A coil of cross sectional area of 4 10-6 m 2 carries a current of 0.2 A. If the magnitude of the magnetic dipole is 12 10-5 A.m 2, calculate the number of the coil s turns. Solution The magnitude of the magnetic moment is 5 12 10 NiA N 150 turns 6 ia 0.2 4 10