Сила, действующая на проводник с током в магнитном поле — Студопедия

Французский физик Доминик Франсуа Араго (1786-1853) на заседании Парижской академии наук рассказал об опытах Эрстеда и повторил их. Араго предложил естественное, как всем казалось, объяснение магнитного действия электрического тока: проводник в результате протекания по нему электрического тока превращается в магнит. На демонстрации присутствовал другой академик, математик Андре Мари Ампер. Он предположил, что суть вновь открытого явления – в движении заряда, и решил сам провести необходимые измерения. Ампер был уверен, что замкнутые токи эквивалентны магнитам. 24 сентября 1820 г. он подключил к вольтову столбу две проволочные спирали, которые превратились в магниты.

Т.о. катушка с током создает такое же поле, что и полосовой магнит. Ампер создал прообраз электромагнита, обнаружив, что стальной брусок, помещенный внутрь спирали с током, намагничивается, многократно усиливая магнитное поле. Ампер предположил, что магнит представляет собой некоторую систему внутренних замкнутых токов и показал (и на основе опытов, и помощью расчетов), что малый круговой ток (виток) эквивалентен маленькому магнитику, расположенному в центре витка перпендикулярно его плоскости, т.о. всякий контур с током можно заменить магнитом бесконечно малой толщины.

Гипотеза Ампера, что внутри любого магнита существуют замкнутые токи, наз. гипотезой о молекулярных токах и легла в основу теории взаимодействия токов – электродинамики.

На проводник с током, находящийся в магнитном поле, действует сила, которая определяется только свойствами поля в том месте, где расположен проводник, и не зависит от того, какая система токов или постоянных магнитов создала поле. Магнитное поле оказывает на рамку с током ориентирующее действие. Следовательно, вращающий момент, испытываемый рамкой, есть результат действия сил на отдельные ее элементы.

Закон Ампера: сила, действующая на элемент длины проводника с током I, помещенного в магнитное поле ,

где сила, — вектор элемента длины проводника, проведенный в направлении тока.

Модуль магнитной силы: , где угол между и .

Следовательно, когда проводник расположен вдоль линий поля , магнитная сила отсутствует.

Направление вектора может быть найдено по общим правилам векторного произведения. В простейшем случае, когда проводник с током и поле взаимно перпендикулярны , для определения направления магнитной силы можно воспользоваться правилом левой руки: если ладонь левой руки расположить так, чтобы в нее входил вектор , а четыре вытянутых пальцев расположить по направлению тока в проводнике, то отогнутый большой палец покажет направление силы, действующей на ток.

, где  число свободных электронов в единице объема проводника (концентрация частиц), — заряд электрона, — скорость упорядоченного движения электронов, — площадь поперечного сечения проводника.

В отличие от кулоновских сил, которые являются центростремительными, сила Ампера не является центральной. Она направлена перпендикулярно к линиям магнитной индукции.

Закон Ампера может быть использован для определения модуля вектора магнитной индукции. Модуль вектора индукции в данной точке однородного магнитного поля равен наибольшей силе, которая действует на помещенный в окрестности данной точки проводник единичной длины, по которому протекает ток в единицу силы тока: . Значение  достигается при условии, что проводник расположен перпендикулярно к линиям индукции.

Закон Ампера применяется для определения силы взаимодействия двух токов.

Между двумя параллельно расположенными бесконечно длинными проводниками, по которым протекают постоянные токи, возникает сила взаимодействия. Проводники с одинаково направленными токами притягиваются, с противоположно направленными токами – отталкиваются.

Сила взаимодействия, приходящаяся на единицу длины каждого из параллельных проводников, пропорциональна величинам токов  и  и обратно пропорциональна расстоянию между R между ними.  Такое взаимодействие проводников с параллельными токами объясняется правилом левой руки. Модуль силы, действующий на два бесконечных прямолинейных тока и , расстояние между которыми равно R:

, т.е..

В неоднородном магнитном поле на контур с током действует сила , где изменение , рассчитанное на единицу длины вдоль направления, совпадающего с направлением .

Сила втягивает магнитный диполь в область больших значений магнитной индукции.

Задача 3. Определить степень неоднородности магнитного поля , если максимальная сила, действующая на точечный магнитный диполь . Магнитный момент точечного диполя =2 мА?м².

Дано: Решение:

,  

90000 11.5: Magnetic Force on a Current-Carrying Conductor 90001 90002 Moving charges experience a force in a magnetic field. If these moving charges are in a wire-that is, if the wire is carrying a current-the wire should also experience a force. However, before we discuss the force exerted on a current by a magnetic field, we first examine the magnetic field generated by an electric current. We are studying two separate effects here that interact closely: A current-carrying wire generates a magnetic field and the magnetic field exerts a force on the current-carrying wire.90003 90004 Magnetic Fields Produced by Electrical Currents 90005 90002 When discussing historical discoveries in magnetism, we mentioned Oersted’s finding that a wire carrying an electrical current caused a nearby compass to deflect. A connection was established that electrical currents produce magnetic fields. (This connection between electricity and magnetism is discussed in more detail in Sources of Magnetic Fields.) 90003 90002 The compass needle near the wire experiences a force that aligns the needle tangent to a circle around the wire.Therefore, a current-carrying wire produces circular loops of magnetic field. To determine the direction of the magnetic field generated from a wire, we use a second right-hand rule. In RHR-2, your thumb points in the direction of the current while your fingers wrap around the wire, pointing in the direction of the magnetic field produced (Figure \ (\ PageIndex {1} \)). If the magnetic field were coming at you or out of the page, we represent this with a dot. If the magnetic field were going into the page, we represent this with an ×.90003 90002 These symbols come from considering a vector arrow: An arrow pointed toward you, from your perspective, would look like a dot or the tip of an arrow. An arrow pointed away from you, from your perspective, would look like a cross or an ×. A composite sketch of the magnetic circles is shown in Figure \ (\ PageIndex {1} \), where the field strength is shown to decrease as you get farther from the wire by loops that are farther separated. 90003 Figure \ (\ PageIndex {1} \): (a) When the wire is in the plane of the paper, the field is perpendicular to the paper.Note the symbols used for the field pointing inward (like the tail of an arrow) and the field pointing outward (like the tip of an arrow). (B) A long and straight wire creates a field with magnetic field lines forming circular loops. 90004 Calculating the Magnetic Force 90005 90002 Electric current is an ordered movement of charge. A current-carrying wire in a magnetic field must therefore experience a force due to the field. To investigate this force, let’s consider the infinitesimal section of wire as shown in Figure \ (\ PageIndex {3} \).The length and cross-sectional area of ​​the section are 90015 dl 90016 and 90015 A 90016, respectively, so its volume is \ (V = A \ cdot dl \). The wire is formed from material that contains 90015 n 90016 charge carriers per unit volume, so the number of charge carriers in the section is \ (nA \ cdot dl \). If the charge carriers move with drift velocity \ (\ vec {v} _d \) the current 90015 I 90016 in the wire is (from Current and Resistance) 90003 90002 \ [I = neAv_d. \] 90003 90002 The magnetic force on any single charge carrier is \ (e \ vec {v} _d \ times \ vec {B} \), so the total magnetic force \ (d \ vec {F} \) on the \ (nA \ cdot dl \) charge carriers in the section of wire is 90003 90002 \ [d \ vec {F} = (nA \ cdot dl) e \ vec {v} _d \ times \ vec {B}.\] 90003 90002 We can define 90015 dl 90016 to be a vector of length 90015 dl 90016 pointing along \ (\ vec {v} _d \), which allows us to rewrite this equation as 90003 90002 \ [d \ vec {F} = neAv_dd \ vec {l} \ times \ vec {B}, \] or 90003 90002 \ [d \ vec {F} = Id \ vec {l} \ times \ vec {B}. \ Label {11.12} \] 90003 90002 This is the magnetic force on the section of wire. Note that it is actually the net force exerted by the field on the charge carriers themselves. The direction of this force is given by RHR-1, where you point your fingers in the direction of the current and curl them toward the field.Your thumb then points in the direction of the force. 90003 Figure \ (\ PageIndex {2} \): An infinitesimal section of current-carrying wire in a magnetic field. 90002 To determine the magnetic force \ (\ vec {F} \) on a wire of arbitrary length and shape, we must integrate Equation \ ref {11.12} over the entire wire. If the wire section happens to be straight and 90015 B 90016 is uniform, the equation differentials become absolute quantities, giving us 90003 90002 \ [\ vec {F} = I \ vec {l} \ times \ vec {B}.\] 90003 90002 This is the force on a straight, current-carrying wire in a uniform magnetic field. 90003 90002 Example \ (\ PageIndex {1} \): Balancing the Gravitational and Magnetic Forces on a Current-Carrying Wire 90003 90002 A wire of length 50 cm and mass 10 g is suspended in a horizontal plane by a pair of flexible leads (Figure \ (\ PageIndex {3} \)). The wire is then subjected to a constant magnetic field of magnitude 0.50 T, which is directed as shown. What are the magnitude and direction of the current in the wire needed to remove the tension in the supporting leads? 90003 Figure \ (\ PageIndex {3} \): (a) A wire suspended in a magnetic field.(B) The free-body diagram for the wire. 90002 90055 Strategy 90056 90003 90002 From the free-body diagram in the figure, the tensions in the supporting leads go to zero when the gravitational and magnetic forces balance each other. Using the RHR-1, we find that the magnetic force points up. We can then determine the current 90015 I 90016 by equating the two forces. 90003 90002 90055 Solution 90056 90003 90002 Equate the two forces of weight and magnetic force on the wire: 90003 90002 \ [mg = IlB.2)} {(0.50 \, m) (0.50 \, T)} = 0.39 \, A. \] 90003 90002 90055 Significance 90056 90003 90002 This large magnetic field creates a significant force on a length of wire to counteract the weight of the wire. 90003 90002 Example \ (\ PageIndex {2} \): Calculating Magnetic Force on a Current-Carrying Wire 90003 90002 A long, rigid wire lying along the 90015 y 90016 -axis carries a 5.0-A current flowing in the positive 90015 y 90016 -direction. (A) If a constant magnetic field of magnitude 0.30 T is directed along the positive 90015 x 90016 -axis, what is the magnetic force per unit length on the wire? (B) If a constant magnetic field of 0.30 T is directed 30 degrees from the + 90015 x 90016 -axis towards the + 90015 y 90016 -axis, what is the magnetic force per unit length on the wire? 90003 90002 90055 Strategy 90056 90003 90002 The magnetic force on a current-carrying wire in a magnetic field is given by \ (\ vec {F} = I \ vec {l} \ times \ vec {B} \). For part a, since the current and magnetic field are perpendicular in this problem, we can simplify the formula to give us the magnitude and find the direction through the RHR-1.The angle 90015 θ 90016 is 90 degrees, which means \ (sin \, \ theta = 1. \) Also, the length can be divided over to the left-hand side to find the force per unit length. For part b, the current times length is written in unit vector notation, as well as the magnetic field. After the cross product is taken, the directionality is evident by the resulting unit vector. 90003 90002 90055 Solution 90056 90003 90104 90105 We start with the general formula for the magnetic force on a wire.We are looking for the force per unit length, so we divide by the length to bring it to the left-hand side. We also set \ (sin \, \ theta \). The solution therefore is \ [F = IlB \, sin \, \ theta \] \ [\ frac {F} {l} = (5.0 \, A) (0.30 \, T) \] \ [\ frac {F} {l} = 1.5 \, N / m. \] Directionality: Point your fingers in the positive 90015 y 90016 -direction and curl your fingers in the positive 90015 x 90016 -direction. Your thumb will point in the \ (- \ vec {k} \) direction. Therefore, with directionality, the solution is \ [\ frac {\ vec {F}} {l} = -1.o) \ hat {i} \] \ [\ vec {F} / l = -1.30 \ hat {k} \, N / m. \] 90110 90113 90002 90055 Significance 90056 90003 90002 This large magnetic field creates a significant force on a small length of wire. As the angle of the magnetic field becomes more closely aligned to the current in the wire, there is less of a force on it, as seen from comparing parts a and b. 90003 90002 Exercise \ (\ PageIndex {1} \) 90003 90002 A straight, flexible length of copper wire is immersed in a magnetic field that is directed into the page.(A) If the wire’s current runs in the + 90015 x 90016 -direction, which way will the wire bend? (B) Which way will the wire bend if the current runs in the — 90015 x 90016 -direction? 90003 90002 90055 Solution 90056 90003 90002 a. bends upward; b. bends downward 90003 90002 Example \ (\ PageIndex {3} \): Force on a Circular Wire 90003 90002 A circular current loop of radius 90015 R 90016 carrying a current 90015 I 90016 is placed in the 90015 xy 90016 -plane. A constant uniform magnetic field cuts through the loop parallel to the 90015 y 90016 -axis (Figure \ (\ PageIndex {4} \)).Find the magnetic force on the upper half of the loop, the lower half of the loop, and the total force on the loop. 90003 Figure \ (\ PageIndex {4} \): A loop of wire carrying a current in a magnetic field. 90002 90055 Strategy 90056 90003 90002 The magnetic force on the upper loop should be written in terms of the differential force acting on each segment of the loop. If we integrate over each differential piece, we solve for the overall force on that section of the loop. The force on the lower loop is found in a similar manner, and the total force is the addition of these two forces.90003 90002 90055 Solution 90056 90003 90002 A differential force on an arbitrary piece of wire located on the upper ring is: 90003 90002 \ [dF = I B \, sin \, \ theta \, dl, \] where \ (\ theta \) is the angle between the magnetic field direction (+ 90015 y 90016) and the segment of wire. A differential segment is located at the same radius, so using an arc-length formula, we have: 90003 90002 \ [dl = Rd \ theta \] 90003 90002 \ [dF = IBR \, sin \, \ theta \, d \ theta. \] 90003 90002 In order to find the force on a segment, we integrate over the upper half of the circle, from 0 to \ (\ pi \).0 sin \, \ theta \, d \ theta = IBR (-cos 0 + cos \ pi) = -2 IBR. \] 90003 90002 The net force is the sum of these forces, which is zero. 90003 90002 90055 Significance 90056 90003 90002 The total force on any closed loop in a uniform magnetic field is zero. Even though each piece of the loop has a force acting on it, the net force on the system is zero. (Note that there is a net torque on the loop, which we consider in the next section.) 90003 .90000 Magnetic Force On a Current-carrying Conductor 90001 90002 Magnetic Force On a Current-carrying Conductor 90003 90004 Because charges ordinarily can not escape a conductor, the magnetic force on charges moving in a conductor is transmitted to the conductor itself. 90005 90006 90006 90004 The magnetic field exerts a force on a current-carrying wire in a direction given by the right hand rule 1 (the same direction as that on the individual moving charges). This force can easily be large enough to move the wire, since typical currents consist of very large numbers of moving charges.90005 90004 We can derive an expression for the magnetic force on a current by taking a sum of the magnetic forces on individual charges. (The forces add because they are in the same direction.) The force on an individual charge moving at the drift velocity \ ({v} _ {d} \) is given by \ (F = {\ text {qv}} _ {d} B \ phantom {\ rule {0.25em} {0ex}} \ text {sin} \ phantom {\ rule {0.25em} {0ex}} \ theta \). Taking \ (B \) to be uniform over a length of wire \ (l \) and zero elsewhere, the total magnetic force on the wire is then \ (F = ({\ text {qv}} _ {d} B \ phantom {\ rule {0.25em} {0ex}} \ text {sin} \ phantom {\ rule {0.25em} {0ex}} \ theta) (N) \), where \ (N \) is the number of charge carriers in the section of wire of length 90011 \ (l \) 90012. 90005 90004 Now, \ (N = \ text {nV} \), where \ (n \) is the number of charge carriers per unit volume and \ (V \) is the volume of wire in the field. Noting that \ (V = \ text {Al} \), where \ (A \) is the cross-sectional area of ​​the wire, then the force on the wire is \ (F = ({\ text {qv}} _ {d} B \ phantom {\ rule {0.25em} {0ex}} \ text {sin} \ phantom {\ rule {0.25em} {0ex}} \ theta) (\ text {nAl}) \).Gathering terms, 90005 90004 \ (F = ({\ text {nqAv}} _ {d}) \ text {lB} \ phantom {\ rule {0.25em} {0ex}} \ text {sin} \ phantom {\ rule {0.25em} {0ex}} \ theta. \) 90005 90004 Because \ ({\ text {nqAv}} _ {d} = I \) (see Current), 90005 90004 \ (F = \ text {IlB} \ phantom {\ rule {0.25em} {0ex}} \ text {sin} \ phantom {\ rule {0.25em} {0ex}} \ theta \) 90005 90004 is the equation for 90011 magnetic force on a length \ (l \ ) of wire carrying a current \ (I \) in a uniform magnetic field \ (B \) 90012, as shown in this figure. If we divide both sides of this expression by \ (l \), we find that the magnetic force per unit length of wire in a uniform field is \ (\ cfrac {F} {l} = \ text {IB} \ phantom { \ rule {0.25em} {0ex}} \ text {sin} \ phantom {\ rule {0.25em} {0ex}} \ theta \). The direction of this force is given by RHR-1, with the thumb in the direction of the current \ (I \). Then, with the fingers in the direction of \ (B \), a perpendicular to the palm points in the direction of \ (F \), as in this figure. 90005 90026 90026 90004 The force on a current-carrying wire in a magnetic field is \ (F = \ text {IlB} \ phantom {\ rule {0.25em} {0ex}} \ text {sin} \ phantom {\ rule { 0.25em} {0ex}} \ theta \). Its direction is given by RHR-1.90005 90030 Example: Calculating Magnetic Force on a Current-Carrying Wire: A Strong Magnetic Field 90031 90004 Calculate the force on the wire shown in this figure, given \ (B = 1 \ text {.} \ Text {50 T} \) , \ (l = 5 \ text {.} \ text {00 cm} \), and \ (I = \ text {20} \ text {.} 0 \ phantom {\ rule {0.25em} {0ex}} \ text {A} \). 90005 90004 90035 Strategy 90036 90005 90004 The force can be found with the given information by using \ (F = \ text {IlB} \ phantom {\ rule {0.25em} {0ex}} \ text {sin} \ phantom {\ rule {0.25em} {0ex}} \ theta \) and noting that the angle \ (\ theta \) between \ (I \) and \ (B \) is \ (\ text {90º} \), so that \ ( \ text {sin} \ phantom {\ rule {0.25em} {0ex}} \ theta = 1 \). 90005 90004 90035 Solution 90036 90005 90004 Entering the given values ​​into \ (F = \ text {IlB} \ phantom {\ rule {0.25em} {0ex}} \ text {sin} \ phantom {\ rule {0.25em} {0ex }} \ theta \) yields 90005 90004 \ (F = \ text {IlB} \ phantom {\ rule {0.25em} {0ex}} \ text {sin} \ phantom {\ rule {0.25em} {0ex}} \ theta = (\ text {20} \ text {.0 A}) (0 \ text {.} \ text {0500 m}) (1 \ text {.} \ text {50 T}) (1) \ text { .} \) 90005 90004 The units for tesla are \ (\ text {1 T} = \ cfrac {N} {A \ cdot m} \); thus, 90005 90004 \ (F = 1 \ text {.} \ text {50 N.} \) 90005 90004 90035 Discussion 90036 90005 90004 This large magnetic field creates a significant force on a small length of wire.90005 90004 Magnetic force on current-carrying conductors is used to convert electric energy to work. (Motors are a prime example-they employ loops of wire and are considered in the next section.) Magnetohydrodynamics (MHD) is the technical name given to a clever application where magnetic force pumps fluids without moving mechanical parts. (See this figure.) 90005 90060 90060 90004 Magnetohydrodynamics. The magnetic force on the current passed through this fluid can be used as a nonmechanical pump.90005 90004 A strong magnetic field is applied across a tube and a current is passed through the fluid at right angles to the field, resulting in a force on the fluid parallel to the tube axis as shown. The absence of moving parts makes this attractive for moving a hot, chemically active substance, such as the liquid sodium employed in some nuclear reactors. Experimental artificial hearts are testing with this technique for pumping blood, perhaps circumventing the adverse effects of mechanical pumps.(Cell membranes, however, are affected by the large fields needed in MHD, delaying its practical application in humans.) 90005 90004 MHD propulsion for nuclear submarines has been proposed, because it could be considerably quieter than conventional propeller drives. The deterrent value of nuclear submarines is based on their ability to hide and survive a first or second nuclear strike. As we slowly disassemble our nuclear weapons arsenals, the submarine branch will be the last to be decommissioned because of this ability (See this figure.) Existing MHD drives are heavy and inefficient-much development work is needed. 90005 90068 90068 90004 An MHD propulsion system in a nuclear submarine could produce significantly less turbulence than propellers and allow it to run more silently. The development of a silent drive submarine was dramatized in the book and the film 90011 The Hunt for Red October 90012. 90005 90002 Summary 90003 90076 90077 The magnetic force on current-carrying conductors is given by 90004 \ (F = \ text {IlB} \ phantom {\ rule {0.25em} {0ex}} \ text {sin} \ phantom {\ rule {0.25em} {0ex}} \ mathrm {\ theta,} \) 90005 90004 where 90011 \ (I \) 90012 is the current, \ (l \) is the length of a straight conductor in a uniform magnetic field 90011 \ (B \) 90012, and 90011 \ (\ theta \) 90012 is the angle between 90011 \ (I \) 90012 and 90011 \ (B \) 90012 . The force follows RHR-1 with the thumb in the direction of 90011 \ (I \) 90012. 90005 90094 90095.90000 Magnetic Field Formula 90001 90002 When electric current is carried in a wire, a magnetic field is formed around it. The magnetic field lines form concentric circles around the wire. The magnetic field direction depends on the direction of the current. It can be determined using the «right hand rule», by pointing the thumb of your right hand in the direction of the current. The direction of the magnetic field lines is the direction of your curled fingers. The magnitude of the magnetic field depends on the amount of current, and the distance from the charge-carrying wire.The formula includes the constant 90003. This is called the permeability of free space, and has a value 90004. The unit of magnetic field is the Tesla, T. 90005 90002 90007 90005 90002 90010 90005 90002 B = magnetic field magnitude (Tesla, T) 90005 90002 90003 = permeability of free space (90016) 90005 90002 I = magnitude of the electric current (Amperes, A) 90005 90002 r = distance (m) 90005 90002 Magnetic Field Formula Questions: 90005 90002 1) What is the magnitude of the magnetic field 0.10 m away from a wire carrying a 3.00 A current? If the current has a vector direction out of the page (or screen), what is the direction of the magnetic field? 90005 90002 Answer: The magnitude of the magnetic field can be calculated using the formula: 90005 90002 90010 90005 90002 90032 90005 90002 90035 90005 90002 90038 90005 90002 90041 90005 90002 90044 90005 90002 90047 90005 90002 90050 90005 90002 90053 90005 90002 90056 90005 90002 The magnitude of the magnetic field is 6.00 x 10 90059 -6 90060 T, which can also be written as 90061 (micro-Tesla). 90005 90002 The direction of the magnetic field can be determined using the «right hand rule», by pointing the thumb of your right hand in the direction of the current. The direction of the magnetic field lines is the direction of your curled fingers. The current has a vector direction out of the page, and so your fingers will curl in the counter-clockwise direction. Therefore, the magnetic field lines point in the counter-clockwise direction, forming circles around the wire.90005 90002 2) If the magnitude of a magnetic field 2.00 m away from a wire is 10.0 nT (nano-Tesla), what is the magnitude of the electric current carried by the wire? If the magnetic field lines form clockwise circles in the plane of the page (or screen), what is the vector direction of the electric current? 90005 90002 Answer: The magnitude of the electric current can be calculated by rearranging the magnetic field formula: 90005 90002 90070 90005 90002 90073 90005 90002 90076 90005 90002 The magnitude of the magnetic field is given in nano-Tesla.The prefix «nano» means 10 90059 -9 90060, and so 90081. The magnitude of the magnetic field at the distance specified is thus: 90005 90002 B = 10.0 nT 90005 90002 90086 90005 90002 90089 90005 90002 90092 90005 90002 90095 90005 90002 The magnitude of the current in the wire is: 90005 90002 90076 90005 90002 90103 90005 90002 90106 90005 90002 90109 90005 90002 90112 90005 90002 90115 90005 90002 90118 90005 90002 The magnitude of the electric current in the wire is 0.100A. 90005 90002 The direction of the electric current can be determined using the «right hand rule». The magnetic field lines form clockwise circles in the plane of the page, so imagine curling your right hand so that your fingers point in the clockwise direction. To do this, your thumb must point toward the page (or screen). Therefore, the direction of the electric current is into the page (or screen). 90005.90000 magnetism | Definition, Examples, Physics, & Facts 90001 90002 90003 Magnetism 90004, phenomenon associated with magnetic fields, which arise from the motion of electric charges. This motion can take many forms. It can be an electric current in a conductor or charged particles moving through space, or it can be the motion of an electron in an atomic orbital. Magnetism is also associated with elementary particles, such as the electron, that have a property called spin. 90005 90006 Fundamentals 90007 90002 Basic to magnetism are magnetic fields and their effects on matter, as, for instance, the deflection of moving charges and torques on other magnetic objects.Evidence for the presence of a magnetic field is the magnetic force on charges moving in that field; the force is at right angles to both the field and the velocity of the charge. This force deflects the particles without changing their speed. The deflection can be observed in the torque on a compass needle that acts to align the needle with the magnetic field of Earth. The needle is a thin piece of iron that has been magnetized-i.e., A small bar magnet. One end of the magnet is called a north pole and the other end a south pole.The force between a north and a south pole is attractive, whereas the force between like poles is repulsive. The magnetic field is sometimes referred to as magnetic induction or magnetic flux density; it is always symbolized by 90009 90003 B 90004 90012. Magnetic fields are measured in units of tesla (T). (Another unit of measure commonly used for 90009 90003 B 90004 90012 is the gauss, though it is no longer considered a standard unit. One gauss equals 10 90017 -4 90018 tesla.) 90005 90002 A fundamental property of a magnetic field is that its flux through any closed surface vanishes.(A closed surface is one that completely surrounds a volume.) This is expressed mathematically by div 90009 90003 B 90004 90012 = 0 and can be understood physically in terms of the field lines representing 90009 90003 B 90004 90012. These lines always close on themselves, so that if they enter a certain volume at some point, they must also leave that volume. In this respect, a magnetic field is quite different from an electric field. Electric field lines can begin and end on a charge, but no equivalent magnetic charge has been found in spite of many searches for so-called magnetic monopoles.90005 90002 The most common source of magnetic fields is the electric current loop. It may be an electric current in a circular conductor or the motion of an orbiting electron in an atom. Associated with both these types of current loops is a magnetic dipole moment, the value of which is 90009 i 90012 90009 A 90012, the product of the current 90009 i 90012 and the area of ​​the loop 90009 A 90012. In addition, electrons, protons, and neutrons in atoms have a magnetic dipole moment associated with their intrinsic spin; such magnetic dipole moments represent another important source of magnetic fields.A particle with a magnetic dipole moment is often referred to as a magnetic dipole. (A magnetic dipole may be thought of as a tiny bar magnet. It has the same magnetic field as such a magnet and behaves the same way in external magnetic fields.) When placed in an external magnetic field, a magnetic dipole can be subjected to a torque that tends to align it with the field; if the external field is not uniform, the dipole also can be subjected to a force. 90005 Get exclusive access to content from our тисяча сімсот шістьдесят вісім First Edition with your subscription.Subscribe today 90002 All matter exhibits magnetic properties to some degree. When placed in an inhomogeneous field, matter is either attracted or repelled in the direction of the gradient of the field. This property is described by the magnetic susceptibility of the matter and depends on the degree of magnetization of the matter in the field. Magnetization depends on the size of the dipole moments of the atoms in a substance and the degree to which the dipole moments are aligned with respect to each other.Certain materials, such as iron, exhibit very strong magnetic properties because of the alignment of the magnetic moments of their atoms within certain small regions called domains. Under normal conditions, the various domains have fields that cancel, but they can be aligned with each other to produce extremely large magnetic fields. Various alloys, like NdFeB (an alloy of neodymium, iron, and boron), keep their domains aligned and are used to make permanent magnets. The strong magnetic field produced by a typical three-millimetre-thick magnet of this material is comparable to an electromagnet made of a copper loop carrying a current of several thousand amperes.In comparison, the current in a typical light bulb is 0.5 ampere. Since aligning the domains of a material produces a magnet, disorganizing the orderly alignment destroys the magnetic properties of the material. Thermal agitation that results from heating a magnet to a high temperature destroys its magnetic properties. 90005 90002 Magnetic fields vary widely in strength. Some representative values ​​are given in the Table. 90005 90044 90045 90046 90047 Typical magnetic fields 90048 90049 90050 90051 90046 90053 inside atomic nuclei 90054 90055 10 90017 11 90018 T 90054 90049 90046 90053 in superconducting solenoids 90054 90055 20 T 90054 90049 90046 90053 in a superconducting coil cyclotron 90054 90055 5 T 90054 90049 90046 90053 near a small ceramic magnet 90054 90055 0.1 T 90054 90049 90046 90053 Earth’s field at the Equator 90054 90055 4 (10 90017 -5 90018) T 90054 90049 90046 90053 in interstellar space 90054 90055 2 (10 90017 -10 90018) T 90054 90049 90094 90095 .