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Transcript

Chapter IV Magnetic Force Magnetic Field: A magnetic field is generated when electric charge carriers such as electrons move through space or within an electrical conductor. The symbol H refers to the strength of magnetic field. Figure (4.1): Magnetic field of an ideal cylindrical magnet with its axis of symmetry inside the image plane. The magnetic field is represented by magnetic field lines, which show the direction of the field at different points. Magnetic Induction: The Magnetic induction (B) is defined as follows: B 0 H Where: 0 is the permittivity and equals to 0 4 xx10 7 wb / A.m H is the magnetic field strength. The magnetic induction (B) is measured in Tesla (T) or in Newton/ Ampere. Meter (N/A. m). Magnetic Flux: The Magnetic flux (φ) is defined as follows: B A Where: B is the magnetic induction and A is the area ℓ H N HN Hs S b a Figure 4.1: Calculation of magnetic field at the point b. The strength of magnetic field (H) at the point b can be expressed as follows: H 1 2 Pm 4 a3 Where: Pm is called the magnetic moment and equals to Pm = m ℓ m is the strength of the magnet ℓ is the length of the magnet Note that: No. Symbol Title Units 1 H Magnetic field strength Ampere/m 2 B Magnetic induction Weber/m2 = Tesla 3 Φ Magnetic flux Weber 4 Permeability Weber/Ampere.m Biot-Savart Law: Biot and Savart arrived at a mathematical I expression that gives the magnetic field at dB some point in space in terms of the current that produces the field. That expression is P based on the following experimental θ observations for the magnetic field d B at a r point P associated with a length element dℓ ℓℓ d of a wire carrying a steady current I. The vector d B is perpendicular both to d (which points in the direction of the current) and to the unit vector r^ directed from d S toward P. The magnitude of d B is inversely proportional to r2, where r is the distance from d to P. The magnitude of d S is proportional to the current I and to the magnitude dℓ of the length element d . The magnitude of d B is proportional to sin θ, where θ is the angle between the vectors d and r^. These observations are summarized in the mathematical expression known today as the Biot–Savart law: dB 0 I d sin 4 r2 Applications on Biot-Savart Law: Magnetic Induction of a Straight Conductor: Consider a thin, straight wire of finite length carrying a constant current I and placed y d =dx P along the x axis as shown in the Figure. Determine the magnitude and direction of r θ the magnetic field at point P due to this a current. x d In this case, the magnetic field (H) can be x O I expressed as follows: H 1 2 1 a and the magnetic induction (B) is expressed as follows: B 0 H 0 1 2 a Example (4.--) A current of 5 A passes through in a long conductor. Calculate the magnetic field strength and magnetic induction at a point 10 cm away from the middle of the wire. Solution: 1- H 1 2 1 1 1 7.96 A / m a 2 x3.14 10 x10 2 2- B 0 H 4 x3.14 x10 7 x 7.96 10 5 Wb / m 2 Magnetic Induction of a Circular Conductor: In this case, the magnetic field (H) in the center of the coil can be expressed as follows: H NI 2r Where, N: number of turns, I is the current r is the radius. and the magnetic induction (B) is expressed as follows: B 0 H 0 NI 2r and the magnetic moment (Pm) is expressed as follows: P m NAI N a 2 I Example (4.--) A current of 3.5 A passes through in a circular coil with number of turns 200 turn and a radius of 20 cm. Calculate the magnetic field strength (H) in the center of the coil, magnetic induction (B) and the magnetic moment (Pm). Solution: The magnetic field (H) H NI 200 x 3.5 2r 2 x 20 x10 2 =1.75x103 A/m The magnetic induction (B) B 0 H 4 x 3.14 x10 7 x 1.75 x103 = 2.2x10-3 Wb/m2 The magnetic moment (Pm) P m NAI N a 2 I 200 x 3.14 x 20 x10 2 2 x3.5 = 88 A. m2 Magnetic Induction of a Solenoid: In this case, the magnetic field (H) in the middle of the coil can be expressed as follows: H NI Also, The magnetic field (H) at the end of the coil can be expressed as follows: H NI 2 Where, N: number of turns, I is the current ℓ is the length of the solenoid. and the magnetic induction (B) is expressed as follows: Example (4.--) A solenoid has 200 turns per centimeter and carrying a current of 1.5 A. Calculate the magnetic field strength and magnetic induction in the middle and at the end of solenoid. Solution: Firstly, The magnetic field (H) in the middle of the solenoid can be expressed as follows: H NI 200 x 1.5 3 x10 4 A.m 2 10 The magnetic induction (B) in the middle of the solenoid can be expressed as follows: B o H 4 x3.14 x10 7 x 3x10 4 3.768 x10 2 Wb / m 2 Secondly, The magnetic field (H) at the end of the solenoid can be expressed as follows: H NI 200 x 1.5 1.5 x10 4 A.m 2 2 x 10 2 The magnetic induction (B) at the end of the solenoid can be expressed as follows: B o H 4 x3.14 x10 7 x 1.5 x10 4 1.884 x10 2 Wb / m 2 Note That: No. Status Magnetic Field (H) 1 H Straight Conductor 1 Magnetic Induction (B) 1 a 2 Circular Conductor 2 NI H 2r 3 Solenoid In the middle At the end B 0 H 0 1 2 a B 0 H 0 H NI In the middle H NI 2 At the end NI 2r B 0 H B 0 H 0 NI 0 NI 2 The Magnetic Force Between Two Parallel Conductors Consider two long, straight, parallel wires separated by a distance a and carrying currents I 1 and I 2 in the same direction as in the Figure 4.3. Let’s determine the force exerted on one wire due to the magnetic field set up by the other wire. Wire 2, which carries a current I2 and is identified arbitrarily as the source wire, creates a magnetic field B 2 at the location of wire 1, the test wire. The magnitude of this magnetic field is the same at all points on wire 1. The direction of B 2 is perpendicular to 1 ℓ wire 1 as shown in the Figure. I1 According to the following equation, B2 FB I LxB 2 F1 a the magnetic force on a length ℓ of wire 1 I2 is F 1 = I1 x B 2 . Figure 4.2 Because ℓ is perpendicular to B 2 , in this situation, the magnitude of F 1 is F 1 = I1 B2 . Because the magnitude of B 2 is given by 0 I 2 a Therefore, B2 I F1 I1 B2 I1 0 2 2a I I F1 0 1 2 2a Example (4.--): A conductor carrying a current of 10 A is placed in a magnetic field with induction of 1.2 Wb/m2. Calculate the force per unit length (F/ℓ) in the following cases: (a) If the direction of wire length is perpendicular to the direction of magnetic field. (b) If the direction of wire length is Inclined at an angle of 00, 300, 600 to the direction of magnetic field. Solution: Note that: F I B sin (a) F 10 x 1.2 x sin 90 12 N / m (b) F 10 x 1.2 x sin 0 0 N / m F 10 x 1.2 x sin 30 6 N / m F 10 x 1.2 x sin 60 10.39 N / m Example (4.--): Two parallel conducting plates, a current of 10 A passes through one of them and a current of 20 A passes through in the other one, calculate the following: The force per unit length (F/ℓ). Define the point at which B1= B2 , if the currents (I1 and I2) in the same directions. Define the point point at which B1= B2 , if the currents (I1 and I2) in opposite directions. I2=20 A I1=10 A Solution: The force per unit length (F/ℓ) F 0 I 1 I 2 2 a 30 cm F 4 x x107 x10 x 20 = 1.33x10-4 N/m 2 x x 30 x10 2 The point at which B1= B2 , if the currents (I1 and I2) in the same directions I1 I2 x 0.3 x 10 20 x 0.3 x 20x 10 x 0.3 x 20x 10x 3 30x 3 x 3 0.1 m 30 The point at which B1= B2 , if the currents (I1 and I2) in opposite directions I1 I2 x 0.3 x 10 20 x 0.3 x 20x 10 x 0.3 x 20x 10x 3 10x 3 x 3 0.3 m 10 Ampere’s Law Ampere’s law states that “ The linear integral of magnetic induction (B) around closed loop equals to the total currents (I) inside this loop multiplying by the permeability (0)”. B cos d o I Applications on Ampere’s Law Note That: No. Status Magnetic Field (H) Magnetic Induction (B) 1 H NI NI H 2 r B 0 H 2 Solenoid Toroid 0 N I NI B 0 H 0 2 r Magnetic Force: The existence of a magnetic field B at some point in space can be determined by measuring the magnetic force FB exerted on an appropriate test particle placed at that point. This process is the same in defining the electric field. If we perform such an experiment by placing a particle with charge q in the magnetic field, it is found the following results that are similar to those for experiments on electric forces: The magnetic force is proportional to the charge q of the particle. The magnetic force on a negative charge is directed opposite to the force on a positive charge moving in the same direction. The magnetic force is proportional to the magnitude of the magnetic field vector B . Also, the following results, which are totally different from those for experiments: on electric forces should taken in our consideration: The magnetic force is proportional to the speed v of the particle. If the velocity vector makes an angle θ with the magnetic field, the magnitude of the magnetic force is proportional to sin θ. When a charged particle moves parallel to the magnetic field vector, the magnetic force on the charge is zero. When a charged particle moves in a direction not parallel to the magnetic field vector, the magnetic force acts in a direction perpendicular to both v and B that is, the magnetic force is perpendicular to the plane formed by v and B . F This relation can be written as follows: F q v x B B The magnitude of the magnetic force on a charged particle is F qvB sin θ v So: FB is zero when v is parallel or antiparallel to B (θ = 00 and 1800) and maximum when v is perpendicular to B (θ = 900). Let’s compare the important differences between the electric and magnetic versions of the particle in a field model: The electric force vector is along the direction of the electric field, whereas the magnetic force vector is perpendicular to the magnetic field. The electric force acts on a charged particle regardless of whether the particle is moving, whereas the magnetic force acts on a charged particle only when the particle is in motion. The electric force does work in displacing a charged particle, whereas the magnetic force associated with a steady magnetic field does no work when a particle is displaced because the force is perpendicular to the displacement of its point of application. The SI unit of magnetic field is the newton per coulomb-meter per second, which is called the Tesla (T): 1T 1 N C .m sec or 1T 1 N A. m Example (4.--): An electron in an old-style television picture tube moves toward the front of the tube with a speed of 8x106 m/s along the x axis. Surrounding the neck of the tube z are coils of wire that create a magnetic field of magnitude 0.025 T, directed at an -e angle of 600 to the x axis and lying in the the electron. Solution: FB = |q|vB sin θ = (1.6x1019 C) (8x106 m/s) (0.025 T)(sin 600) = 2.8x 10-14 N y v xy plane. Calculate the magnetic force on x F 600 B Motion of a Charged Particle in a Uniform Magnetic Field The particle moves in a circle because the magnetic force F B is perpendicular to v and B and has a constant magnitude qvB. r q FB As Figure 4.2 illustrates, the rotation is v + counterclockwise for a positive charge in + FB a magnetic field directed into the page. If q FB v q were negative, the rotation would be + clockwise. We use the particle under a net v q force model to write Newton’s second law for the particle: F F B ma Because the particle moves in a circle, we also model it as a particle in uniform circular motion and we replace the acceleration with centripetal acceleration: FB qvB mv 2 r This expression leads to the following equation for the radius of the circular path: r mv qB That is, the radius of the path is proportional to the linear momentum mv of the particle and inversely proportional to the magnitude of the charge on the particle and to the magnitude of the magnetic field. The angular speed of the particle is expressed as follow: v qB r m The period of the motion (the time interval the particle requires to complete one revolution) is equal to the circumference of the circle divided by the speed of the particle: T 2 2r 2m v qB These results show that the angular speed of the particle and the period of the circular motion do not depend on the speed of the particle or on the radius of the orbit. The angular speed v is often referred to as the cyclotron frequency because charged particles circulate at this angular frequency in the type of accelerator called a cyclotron. Example (4.--): A proton is moving in a circular orbit of radius 14 cm in a uniform 0.35-T magnetic field perpendicular to the velocity of the proton. Find the speed of the proton. Solution: v qBr mP Substitute numerical values: q B r 1.6 x10 19 x 0.35 x 0.14 v mP 1.67 x10 27 1.67 3 10227 kg V= 4.7x106 m/s Example (4.--): An electron moves in circular path with radius of 2x10-5 cm under the effect of magnetic field with induction of 1.5x10-3 Wb/m2. Calculate the following: Qe= 1.6x10-19 C and me= 9.11x10-31 kg Solution: The magnetic force (F) F Q v B 1.6 x1019 x 52.68 x1.5x103 1.26 x1020 N The electron velocity (v) eBr 1.6 x1019 x1.5 x103 x 2 x105 x102 m 9.11x10 31 v 52.68 m / sec v The angular velocity of electron (ω) eB 1.6 x1019 x1.5 x103 m 9.11x10 31 263x109 rad / sec . Electromagnetic Induction Introduction What is the physical basis for the static or noise produced in a radio when certain appliances are used in its vicinity? What is the principle supporting the operation of the ferrite rod antenna in a radio? How does the power company generate AC voltage? The answers to all of these questions involve electromagnetic induction. In this part we will study the physical phenomena associated with electromagnetic induction. Induced EMF: Faraday's Law Let us consider a simple experiment. The equipment needed consists of a coil of wire, a galvanometer and a bar magnet. The coil of wire is connected to the galvanometer. We observe the impulsive deflection of the galvanometer under a series of conditions. First, If we move the magnet towards the coil of wire, we see a deflection of the galvanometer. The deflection means that there is a current in the galvanometer coil circuit, hence an induced emf. Second, If we move the magnet away from the coil, we see another deflection of the galvanometer. We observe in these cases that the deflections are in opposite directions. Third, If we reverse the ends of the magnet and repeat the experiment described above, we observe that the galvanometer deflection for approach is opposite to that in the first case and likewise on taking the magnet away. What conclusions can we reach from this series of experiments? Now let us try a set of experiments in which the magnet remains fixed and the coil is moved toward and away from the magnet - first toward one end of the magnet and then toward the other end. We will find that when the coil and a given end of the magnet are approaching each other the direction of deflection is the same regardless of which is moving. Also the direction of deflection is the same when the coil and a given end are receding from each other, regardless of which is moving. We can get additional information relative to the time involved in the process. If we move either the magnet or the coil faster, the deflection is increased. The results of these experiments indicate that an emf is induced in a coil of wire whenever there is a change in the magnetic flux interlinking it, and that there is a definite relationship between the direction of the field and the direction of motion and between the magnitude of the induced emf and how rapidly we change the magnetic flux. On the basis of experiments such as these, Michael Faraday, one of the greatest experimental investigators in the history of science, formulated his now famous law for electromagnetic induction. Faraday envisioned the magnetic field associated with a magnet to be ethereal tentacles that grasp the iron filings and pull them toward the magnet. His induction law expressed the following idea: When the number of tentacles cutting through the area capping a circuit changes in time, there will be an induced electric field in the circuit and thus an induced voltage and current. We can put this induction law in a useful quantitative form. We define the magnetic flux φ B A B as the magnetic induction field magnitude B, times the area of the circuit that is perpendicular to B. The magnetic flux φ can be written as r B into Paper r 2 B B A cos Where, θ is the angle between the magnetic induction B and the line perpendicular to the area A. Faraday's law states that whenever the magnetic flux through an area changes, there will be an emf induced in the perimeter of the area (or equivalently there will be an electric field induced at the perimeter of the area). We can express this in the following equation: d dt (Weber / sec) Where is the electromotive force in volts and dφ is the change of magnetic flux in webers in a change of time dt in seconds. The negative sign in the equation is due to Lenz's law which states that the induced voltage is in such a direction as to oppose its cause; that is, the magnetic field due to the induced current opposes the changing magnetic flux. The induced current has an associated magnetic flux which tries to maintain the status quo. Applications of Faraday's Law There are many different ways to produce induced currents. Faraday's law indicates that a changing field B, a changing area, or a changing angle between B and the circuit can each produce an induced emf. We now consider some specific quantitative examples. Consider a U-shaped conductor in a plane perpendicular to a magnetic field with induction B. A rod of length ℓ is moving to the right with a velocity v as shown in the Figure. + According to Faraday's law: d Bd dt t I B v where d v - B is the magnitude of the magnetic induction in tesla, ℓ is the length of the rod, d is the distance in meters through which the rod moves in t seconds, and v is the magnitude of the velocity of the rod A Simple generator. When the rod is moved to the right as shown an emf will be induced across the length Blv. The top end of the rod will be positive Example (4.--) Find the induced emf in a 0.50 m rod moving with velocity of 2 m/sec perpendicular to a field of 2 x 10-4 Wb/m2, or tesla. Solution: B v = 2 x 10-4x 2 x 0.50 = 2 x 10-4 Volt Suppose a coil of wire of 1cm radius is removed from a field of 0.1 Wb/min 0.01 sec. The induced emf in the coil will be: BA , t where, A d , then 0.1 x 3.14 x 0.01 = 10 2 2 3.14 x10 3 Volt 3.14 mVolt If the coil has 100 turns of wire, the flux is increased by a factor of 100: φ= NBA = 100 BA Such a coil can be used as a search coil to measure an unknown magnet Self Inductance A single coil of many turns has inductive properties due to the magnetic interaction of adjacent loops in the coil. If the loops are wound in the same direction the total effect of this interaction is an induced back emf that opposes the initial current change. This self-inductance is a form of inertia resisting current change in the coil. The selfinduced emf can be expressed in equation form as follows: L where di dt L is the self-inductance (di/dt) is the current change per unit time in the coil. If is measured in volts and (di/dt) in amperes per second, the inductance is given in Henries, 1 H = 1 V sec/A. The negative sign is consistent with Lenz's law and conservation of energy. Example (4.--) If the current changes in an air core inductor having a self-inductance of 25 x 10-5 H from zero to 1.0 A in 0.10 sec, what is the magnitude and direction of the self induced emf? Solution: L di 1 2.5 X 10 3 Volt = 25 x10 5 x dt 0.1 As the current is increasing, the direction of the induced emf opposes the growing current; hence, the negative sign. The effect of an inductor in a circuit is to oppose any change in current. Thus, inductance is analogous to inertia in a mechanical system. To minimize self-inductance in wire-wound resistors, the loops are wound back on themselves. This non-inductive winding tends to cancel out the inductive interactions of the loops. Inductors An inductor is a device that stores energy in the magnetic field set up by the current through a coil. By analogy the inductor plays the same role for magnetic field energy that the capacitor does for the electric field energy. We can derive an equation for the energy stored in an inductor as follows: the work done in moving charge q against the back emf is W q L di i dt , dt where q i dt If we start with the current at zero and the current increases at a constant rate to the final value I, we can use the average value of I/2 to get the work done. We then get the following equation for the work done in building up the magnetic field: W 1 LI2 2 = energy stored in the field This represents the work done in building up a current of I in the self-inductor and establishing the magnetic field. It also represents the energy that is stored in the magnetic field and that becomes available if the current is reduced to zero. This is shown by the flash on opening a switch in a DC circuit in which there is an inductor. Example (4.--) The back emf induced in a coil when the current changes from 100 mA to zero in 1 m. sec is 1 Volt. Find the self-inductance of the coil and the energy dissipated when the current goes to zero. Solution: The self inductance of the coil is L di dt 100x10 1 3 1x10 3 0.01 Henry The energy dissipated when the current goes to zero is W 2 1 1 L I 2 x 0.01 x 100 x10 3 0.5 x10 4 J 2 2 Combinations of Inductors (a) Series connection (b) Parallel connection I I L1 V L2 V L1 L2 L3 I1 I2 I3 L3 When the inductors are connected in series (Figure a) The current is the same through all of the inductors and the potential is the sum of the individual emfs, V 1 2 3 V L di di1 di L 2L 3 dt dt dt V Leff di dt Where L1, L2 and L3 are the inductances and where dI is the change in the current in the system in a time dt. For a series configuration, all of the currents are the same: di1 di2 di3 di V L1 L2 L3 Thus, di dt Leff di dt Leff L1 L2 L3 When inductors are connected in series, their effective inductance is the sum of the individual inductances. When the inductors are connected in parallel (Figure b), The potential across all of the inductors is the same, but the current through the system is divided into components so that the total current is the sum of the individual components, I di1 di2 di3 and V L1 di di di1 di L2 2 L3 3 Leff dt dt dt dt Then solving for, di1/dt, di2/dt and di3/dt, we have: di di di1 di 2 3 dt dt dt dt V V V Leff L1 L2 V L3 Thus 1 1 1 1 Leff L1 L2 L3 When inductors are connected in parallel, their effective inductance is the reciprocal of the sum of the reciprocals of the individual inductance. Example (4.---) Inductors of 3 H, 5 H, and 10 H are connected a. in series b. in parallel c. with the 3H inductor in series with a parallel connection of the 5H and 10 H inductors. What is the effective inductance of each configuration? Solution: (a) Series Leff L1 L2 L3 = 3 H + 5 H + 10 H = 18 H (b) Parallel Parallel 1 1 1 1 Leff L1 L2 L3 1 1 1 1 19 Leff 3 5 10 30 Leff 1.6 H (c ) Combination 5 x10 1 1 Leff 3 3 5 10 5 10 Leff 6.3 H