AAST/AEDT 1 AP PHYSICS-C: MAGNETIC FIELD Let us run an experiment. We place two parallel wires close to each other. If we turn the current on, the wires start to interact. If currents are opposite by their direction then the repulsion is observed. Currents attract if the directions are equal. If the current exists only in one wire, then there is no interaction. We also have no interaction if the current interacts with another current in a twisted wires. The conclusion from the observations is: Current changes the properties of the surrounding space. That change creates a force that is exerted on any other current placed into this space. The space with the property to exert a force on the current we define as a MAGNETIC FIELD. As you can see there is an analogy between electric and magnetic fields There are no magnetic charges and so we can not use them to study the magnetic field. That is why magnetic field is studied with a small loop or frame of wire with a current. That loop must be hanged by the elastic wires. When the current is flowing though the loop the magnetic forces create a torque. The greater the torgue the more the angle of
2 rotation. By measuring the magnitude of that angle we can estimate the strength of the magnetic field. The same loop can be used to discover the direction of the field. As we studied before in mechanics, a torque is the product of the applied force and lever arm. On a diagram (a) the lever arm is nonzero, so the torgue 0 and the loop rotates. If the connected twisted wires are nonelastic (the loop rotates freely), then loop will rotate until its lever arm and consequently the torgue will become zero ( b). Scientists agreed to define the direction of magnetic field as the direction of the perpendicular to the surface of such stopped loop. That loop should be initially suspended on nonelastic twisted wires. Because every loop has two perpendiculars the corkscrew rule had been established. If the corkscrew s handle rotates as current directed than the direction of its motion will coincide with the direction of the magnetic field. Another way to discover the direction of the field is to use a compass or a magnetic arrow. If we place it into the field, it starts to rotate and finally comes to rest. Then the direction from the S pole toward the N pole is the direction of the magnetic field.
3 If we place different loops with different currents at the same point of the magnetic field and measure the maximum torque applied on them, we will observe that the ratio of the torque over the current vs. area product is equal for each loop. That means that that ratio depends only on the field properties and it can be used as the characteristic of the magnetic field. (Analogous to the electric field intensity). That ratio is defined as magnetic induction. Magnetic induction is the ratio of the maximum torque exerted on a loop with current at a given point of the magnetic field over the product of current times loop s area. Where B is the magnetic induction, τ - torque, exerted on a loop I - the current, and A is the area of the cross-section of the loop. As you can observe from the formula B is a vector. Its direction would be detemined below. The unit of the magnetic field induction is Tecla(T) - it is the induction of the magnetic field, where a torque of 1N * m is exerted on a loop with the current of 1 A and with the cross-section of 1 m 2. Another unit that is widely used in mesuring the magnetic induction is Gauss. Gauss was the unit of magnetic indugtion in CGS system units, a system based on cantimeter, gram and second. That system was in use before scientists decided to use SI system. 1 Gauss = 10-4 Tecla. In problems gauss should be changed to Teclas. MAGNETIC FIELD LINES These lines do not exist in nature. They were invented by physicists to visualize magnetic fields. By the definition the magnetic field line is the imaginary line, the tangent of which at any point is always directed as the induction of a magnetic field at the same point. We can observe those lines with iron filings. One examples of magnetic field, you can see below.
4 The field of the bar magnet. More examples to follow BIO-SAVART LAW The goal of Bio and Savart was to study the magnetic field created by an element of current. Let us imagine that we have a wire with a current I and we define an element of length as ds. We also measure the field induction produced by this elementary current in point A, located at a distance described by a vector r from the current element as it shown on a diagram. As the result of their experimental study Bio and Savart discovered that field induction at A is proportional to the elementary current Ids and inversely proportionally to the distance square. The field also proportional to the sin of the angle between the directions of vectors ds and r. Mathematically the Bio-Savart law is In A vector shape the same law can be presented through the vector product of vectors ds and r. µ o / 4π - is a proportionality coefficient, where µ o - is defined as permeability or magnetic constant. It is one of the fundamental constants in nature. Its value is 1.26 10-6 H/m. Direction of db is perpendicular to the plane of ds and r. Knowing the field created by elementary current we can calculate the field created by various conductors. Several examples are presented below. FIELD OF A STRAIGHT CURRENT Let us assume that current I is flowing through an infinite straight wire and we want to determine field induction at point A as it shown on a diagram. The field db created at point A by the current element I ds can be estimated by Bio-Savart s law as To detemine the total field created by the net wire we have to
summarize fields created by each current element, i.e take an integral along the wire with a limits from - to + 5 sinθ can be prezented as sin (π-θ) = R /r and ). Thus The value of the integral can be determine in a table of integrals. The result is Our next job is to resolve uncertainty, caused by the infinity in the limits. To do that first we take limits from (-s) to (s) and than substitute instead of s. The result is Now if we subsitute instead of s, the final result for the field induction created by an infinite straight current would be It is obvious from the formula that magnetic field lines for the field should be a concentric circles, as it shown on a diagram The direction of the magnetic field created by the straight current can be determined by using two similar rules.
Method 1: The corkscrew rule: If the corkscrew moves the same direction as the current, than the direction of the handles rotation coincides with the direction of the magnetic field If you want to know the direction of the corkscrew motion you can use a rule RIghty Tighty, Lefty Loosey. (this rule was suggested to me by one of the students) 6 Method 2: The Right hand rule. If the thumb of your right hand points the direction of current in a straight wire, than your bent fingers point the direction of the magnetic field lines. FIELD OF A SINGLE LOOP WITH CURRENT Let us assume that a field is created by a current I in a single loop with a radius of R. We would like to determine a field at any point on the axis of simmetry for the loop. Let us assume that the point is located at the height Z above the loops plane. Than the elementary current Ids would produce elementary field db as it shown on a diagram. Magnetic field is perpendicular to the plain of vectors r and ds (gray area on a digram). However from the point of view of symmetry it is obvious the we have to summarize only perpendicular components of db. Parallel components would be canceled by the cymmetrical current element on the other side of the loop. According to Bio-Savarts law We have to take into account that the angle between vectors ds and r is 90 and thus
sin θ =1. Cos α can be expressd as R /r. by the loop Thus the total field created 7 In a special case for Z=0, the field in the center of the loop AMPERE S LAW As you recall there is a relationship between the total flux of electric field intensity through the closed surface and the net charge inside it (Gauss s law). Similar law was discovered for marnetic field. There are several differences. *The surface integral for the electric field flux is replaced by the linear integral along the closed line for magnetic field induction. *The total charge inside the surface is replaced by the total current through the closed contour. * Electric constant is replaced by the magnetic one. Thus, the law that establishes general relationship between the induction and the current is. Although the law had been discovered by J C MAxwell the law is called Ampere s Law. In general the law allows us to calculate magnetic field for any current distribution. This can be done with numerical calculations done with a very powerfull computers.we can apply it to calculate magnetic field induction in several symmetrical examples. FIELD OF A STRAIGHT CURRENT This time we will derive its expression with Ampere s law. We pick up a closed contour in a shape of a circle that encloses a current of I (Amphere s loop). Because of a symmetry B should be constant along the contour and we can move it out of the integral. An angle beween vectors B and ds is zero and cosθ in a scalar product is equal to 1. Thus, or
8 The same formula that was derived before. This time the job was done much easier. MAGNETIC FIELD OF A SOLENOID By definition solenoid is a coil of wire. If we drop iron filings on card board through solenoid and turn on the current, we determine the shape of solenoid s magnetic field as it shown on a diagram The field of the coil of wire(solenoid) with a current. It is interesting to observe that the shape of the magnetic fields for a bar magnet and a coil of wire are the same. We will use that similarity later. The field is strong inside the solenoid and much weaker outside of it. Our goal is to determine the magnetic field induction inside a solenoid. The cross-sectiom of the solenoid s field is shown on a diagram. + means that the current is going out of the cross-section, - means that the current is going into. We pick up Amphere s loop in a shape of a rectangle abcd. Let us apply Amphere s law to the infinite loop. If we travel counterclockwise the angle between B and ds is 0 inside the loop on ad and outside on cb and 90 on dc and ba. Where N is the total number of turns along ad If we define ad as L and the ratio of N /L as n - turn s density, i.e. number of wire s turns per unit of length, then magnetic field inside solenoid can be expressed as
MAGNETIC FIELD OF A TOROID Toroid is a donut shaped coil. It is widely used to create strong magnetic field. One of the largest toroids on the planet is located in Princeton Plazma Research Lab in New Jersey. It is used to create a field strong enough to create condensed plasma necessary for the development of governed thermonuclear reaction. Our goal is to derive formula for the field induction inside the toroid. The shape of the field can be determined with iron filings and it shows that the field is located inside the toroid and te outside field is negligible. The cross-section of the toroid s field is shown on a diagram. We also pick up Ampere s loop in a shape of a circle inside the toroid. The angle between the loop element and B is 0. Let R be the radius of Ampere s loop and N the total number of turns. According to Ampere s law Thus, the formula for the toroid s field is 9 or We define, and the final formula is Forces on a current in a magnetic field If we place a conductor with current into magnetic field a force would be exerted on the current. Ampere discovered that the force is equal to where I- is the current, B is the magnetic induction, L-is the wire s length and α is an angle between the directions of the magnetic induction and the current. The direction of the force can be found by using a left-hand rule. Point the fingers in the direction of the current (conventional current), position your palm so, that the magnetic lines are directed into it. Then the unbended thumb would give you the direction of the force acting on the wire. Example:
10 In several text-books you can discover a different right hand rule. You have to orient your right hand so that the outstretched fingers point in the direction of the conventional current. If you bend your four fingers they should be directed as a magnetic field. Then the unbended thumb would give you the direction of the force acting on the wire. Which rule you prefer, it is your option. The phenomenon of the existence of the magnetic forces are widely used in various electrical devices, for example in galvanometers and electric motors. Example: Let us place a rectangular loop in a magnetic field created by two permanent magnets. If we apply the left-hand rule we discover that the downward force is exerted on wire AB and upward force is exerted on wire CD. As for the wires BC and DA - there is no force exerted, because current in those wires is parallel to the direction of the magnetic field. (α = 0, sin α=0 and thus the force = 0). INTERACTION OF THE PARALLEL CURRENTS Let assume that we have two long parallel wires with a currents of I1 and I2 respectively located at a distance of R and we want to estimate the force between them.
11 We should assume that the second wire is located in a magnetic field created by the first one. According to the Biot- Savart law the field created by the current 1 at a distance R is According to the Amphere s formula the force exerted on the current in a magnetic field is If we substitute B into F, the formula for the force between two currents will be Motion of the Single Charged Particle in Magnetic Field. The force exerted on a current could be represented as the sum of the forces exerted on all charged particles, that created those current. Then the force exerted on a single particle has to be N times less then the force exerted on the current, where N is the number of the current carrying particles inside the wire. As we know the current can be expressed as where q- is the charge of a single particle, v is the average particle s velocity, A- the wire s cross-section area, n the particle s density, i.e. the number of the particles per unit volume. If we substitute the expression into expression for the force and also assume that sinα=1 (the particle is moving perpendicular to the magnetic field), we have It is easy to understand, that AL =V, where V- is the volume of the wire, and that the nv=n. That is why we can eliminate nal from the top and N from the bottom. Finally we have This is the expression for the force exerted by the magnetic field on a single charged particle that is moving inside that field. That force is often called the Lorentz force Just as for the force exerted on a current, the direction of the force exerted on a single charged particle can be found by the left-hand rule. The only difference is that instead of,
12 directing fingers along the current, we have to direct them along the direction of motion for positively charged particles and against the direction of the velocity for the negatively charged particles. The unbended thumb is always perpendicular to the fingers. That means that the direction of the force always perpendicular to the velocity. The force that is perpendicular to the velocity creates the centripetal acceleration. So a particle will have a circular trajectory. We can easily find the radius and the period of rotation. According to the second Newton s law F is the force exerted on a particle, a is the centripetal acceleration = V 2 /R, where R is the radius of the rotation. So, we have The period can be computed as 2πR/v - the circumference over the velocity. We have As we can see the period does not depend on the velocity. The property is used in accelerators of elementary particles. Home assignment 1: Chapter 28, Ditto Page 757 Problems: #2, 3, 4, 7, 6, 9,11, 15, 18, 24(6ed), 33, 35, 37,39 Questions 4,6,8 Home assignment 2: Chapter 29 Ditto Page 781 Questions: 2,6,8 Problems # 1, 79, 6, 13,12(6ed),17, 30(6ed), 25, 28, 30, 33,34, 36, 37, 40,44
13 PAGE 682 1a 0.0 N 1b 975163307.59 m/s2 1c the same 4a 6.24e-14k N 4b -6.24e-14k N 5a east 5b m/s^2 5c 0.00297569124 m 6271489779890 6a 0.0 N direction k 6b 0.0 N direction K 6c 0.0 N 9 0.00026678057 T 14 0.38235294118 m/s 15 0.00002114821 T 20a 2602409.6385 m/s 20b 0.00000010859 s 20c 140530.12048 ev 20c 70265.060241 V 24 33 d/ 28.190778624 l h t(2) N 35 0.46700879765 A 37 0.1176 T if field is 37 angle for the h i t l than fi ld i 0.1T B /(IL*( 39 0.00433012702 Nm
Questions, Chapter 29 page 681 14 #1 b #2 a,b,c tie = sqrt(64+144) #3 a)electric b)magnetic #4 2, 5, 6, 9,10 #5 a)negative, b)same c)same d)half circle with less R #6 a)into b)out c)out d)into e)out f)out #7 a)-1 b)1-into, 2 -out c)less #8 Let R =mv/qb, then R1= 2R (i), R2=.5R (e) R3= R (c), R4= 3R (a), R5= 4R (g) R6= -2R (j), R7=.25R (d) R8= -R (b), R9= -8R (none?), R10 = -4R (f) R11 =infinity (k) #9 a) 1-180, 2-270, 3-90, 4-0, b) 1 and 2 tie, 3 and 4 tie
15 Questions, chapter 30 #1. c, d, b and a tie #2 a) into, b) greater #3 c, a, b #4 b, d, c and a tie #5 1,3, 2, b) less #6 a, b and d tie, c chapter 30 1 3.29E-6 T yes 4a 4.01E-3 m (one point) 7 angle=2 radians 10a 0 10b µi/4r 10c µi/4r 12 prove 16 29 3.21E-3 N 34a 30 5µi only 7, 6, 3-2µ d 1 34b 13µ 37 prove 40 5.71E-3 T 46 0.27 A #7 c and d tie, b, a #8 b, a, d, c #9 d, a and e tie, b,c #10 a) 2 and -4 b)2, 4 and -6 c) 1, 5, -3 and -6 d) 1, 5, -2, -3, and -4
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