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## Electricity and Magnetism

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**Electricity and Magnetism**Chapter 23**Hans Christian Oersted**• 1777 – 1851 • Best known for observing that a compass needle deflects when placed near a wire carrying a current • First evidence of a connection between electric and magnetic phenomena**Magnetic Fields – Long Straight Wire**• A current-carrying wire produces a magnetic field • The compass needle deflects in directions tangent to the circle • The compass needle points in the direction of the magnetic field produced by the current**Direction of the Field of a Long Straight Wire**• Right Hand Rule #1 • Grasp the wire in your right hand • Point your thumb in the direction of the current • Your fingers will curl in the direction of the field**Magnitude of the Field of a Long Straight Wire**• A straight current-carrying wire generates a cylindrical magnetic field in the space surrounding it. • The magnitude of the field at a distance r from a wire carrying a current of I is • µo = 4 x 10-7 T.m / A • µo is called the permeability of free space • The magnetic field of a long straight wire in a vacuum is**Example 1**• The overhead power cable for a street trolley is strong horizontally 10 m above the ground. A long straight section of it carries 100 amps dc due west. Describe the magnetic field produced by the current, and determine its value at ground level just under the wire. Compare that to the strength of the Earth's field. • Given: r = 10 m and I = 100 A • Find: B**Example 1**• Solution: We've got a straight current-carrying wire and that produces a known B-field. • See diagram on last slide with the current assumed heading west. At ground level (at a point beneath the westerly current), the Right-Hand-Current Rule tells us that B points due south. We already have an expression for the field of a long straight current-carrying wire in terms of I and r. • m0= 4p x 10-7 T-m/A. • Which is only 4% of the Earth's field.**Force Between Two Conductors**• The magnetic fields created by moving charges can interact and create forces much like the electric forces between those charges. • Parallel conductors carrying currents in the same direction attract each other • Parallel conductors carrying currents in the opposite directions repel each other**Magnetic Force Between Two Parallel Conductors**• The force on wire 1 is due to the current in wire 1 and the magnetic field produced by wire 2 • The force per unit length is:**Example 2**• What is the force per unit length experienced by each of two extremely long parallel wires carrying equal 1.0-A currents in opposite directions while separated by a distance of 1 m in vacuum? • Given: I1 = I2 = 1.0 A and d = 1 m • Find: FM**Example 2**• Make a drawing. The defining equation is FM/l = 2 x 10-7N/m, repulsion**André-Marie Ampère**• 1775 – 1836 • Credited with the discovery of electromagnetism • Relationship between electric currents and magnetic fields • Mathematical genius evident by age 12**Ampère’s Law**• André-Marie Ampère found a procedure for deriving the relationship between the current in an arbitrarily shaped wire and the magnetic field produced by the wire • Ampère’s Circuital Law • Sum over the closed path ***THE MORE CURRENT….THE STRONGER THE MAGNETIC FIELD*****Magnetic Field of a Current Loop**• The strength of a magnetic field produced by a wire can be enhanced by forming the wire into a loop • All the segments, ∆x, contribute to the field, increasing its strength**Magnetic Field of a Current Loop – Equation**• The magnitude of the magnetic field at the center of a circular loop with a radius R and carrying current I is • With N loops in the coil, this becomes**Magnetic Field of a Solenoid**• If a long straight wire is bent into a coil of several closely spaced loops, the resulting device is called a solenoid • It is also known as an electromagnet since it acts like a magnet only when it carries a current**Magnetic Field of a Solenoid, 2**• The field lines inside the solenoid are nearly parallel, uniformly spaced, and close together • This indicates that the field inside the solenoid is nearly uniform and strong • The exterior field is nonuniform, much weaker, and in the opposite direction to the field inside the solenoid**Magnetic Field in a Solenoid, 3**• The field lines of the solenoid resemble those of a bar magnet**Example 3**• A 20-cm-long solenoid with a 2.0-cm inside diameter is tightly wound on a hollow quartz cylinder. There are several layers with a total of 20 x 103 turns per meter of a niobium-tin wire. The device is cooled below its critical temperature and becomes superconducting. Since the wire is then without resistance, it can easily carry 30 A and not develop any I2R losses. Compute the approximate field inside the solenoid near the middle. What is its value at either end? • Given: n = 20 x 103 m-1, I = 30 A, and D = 2.0 cm • Find: Bz**Example 3**• Solution: We've got a current-carrying solenoid and that produces a known B-field. • The solenoid is long and narrow and will obey the approximations that led to the last equation • Using m0 = 1-257 X 10-6 T-m/A, Bz ≈ m0nI =(1.257X 10-6 T-m/A)(20 x 103 m-1)(30A) Bz ≈ 0.75 T • Which is a formidable field, over 104 times that of the Earth. The field at either end is about half this, 0.38 T**Moving Charges and Magnetism**• Moving charge creates B • The orientation (direction) of B depends upon the orientation (direction) of I • WITHIN atoms, electrons move in different (often opposing) directions, thus individual B usually cancel out • BETWEEN atoms that DO have B fields, those FIELDS may oppose each other and cancel out **Only in materials that have BOTH alignments do we see magnetic properties*** FERROMAGNETIC**Right Hand Rule #2**• Place your fingers in the direction of • Curl the fingers in the direction of the magnetic field, • Your thumb points in the direction of the force, , on a positive charge • If the charge is negative, the force is opposite that determined by the right hand rule • Maximum force is**Finding the Direction of Magnetic Force**• Experiments show that the direction of the magnetic force is always perpendicular to both and which form a plane. • Fmax occurs when is perpendicular to • F = 0 when is parallel to**Example 4**• A conventional water-cooled electromagnet produces a 3.0-T uniform magnetic field in the 4-in. gap between its flat pole pieces. The field is aligned horizontally pointing due north. A proton is fired into the field region at a speed of 5.0 x 106 m/s. It enters traveling in a vertical north-south plane, heading north and downward at 30° below the horizontal. Compute the force vector acting on the proton at the moment it enters the field. • Given: A proton with v = 5.0 X 106 m/s, at 30° below the horizontal in the northerly direction, and B = 3.0 T, north • Find: FM**Example 4**• Solution: Here a charged particle is moving in a B-field and that should call to mind vxB. • First, make a drawing • The proton has a positive charge of +1.60 X 10-19 C and so v x B is due east, FM is due east • The basic force-on-a-moving-charge relationship is FM= qvB sin q • The angle between v and B is q = 30° and so with q = qe FM = qe vB sin q FM = (+1.6 x 10-19 C)(5.0 x 106 m/s)(3.0 T)(sin 30°) Fm= 1.2 x 10-12 N**Force on a Charged Particle in a Magnetic Field**• Consider a particle moving in an external magnetic field so that its velocity is perpendicular to the field • The force is always directed toward the center of the circular path • The magnetic force causes a centripetal acceleration, changing the direction of the velocity of the particle**Force on a Charged Particle**• Charged particle entering perpendicular to uniform magnetic field and experiences centripetal acceleration • Equating the magnetic and centripetal forces: • Solving for R: • R is proportional to the momentum of the particle and inversely proportional to the magnetic field • Sometimes called the cyclotron equation**Particle Moving in an External Magnetic Field**• If the particle’s velocity is not perpendicular to the field, the path followed by the particle is a spiral • The spiral path is called a helix**Electromagnets**• Electromagnets are created by an electric current travelling through a solenoid. As such, their strength and direction can be controlled. • Strength can be increased by: • Increasing current (I) • Increasing the number of coils (N) – although this means a longer total wire length and thus more resistance. • Increasing the core material’s permeability.**Magnetic Force on Current Carrying Conductor**• A force is exerted on a current-carrying wire placed in a magnetic field • The current is a collection of many charged particles in motion • The direction of the force is given by right hand rule #2**Force on a Wire**• The blue x’s indicate the magnetic field is directed into the page • The x represents the tail of the arrow • Blue dots would be used to represent the field directed out of the page • The • represents the head of the arrow • In this case, there is no current, so there is no force**Force on a Wire**• B is into the page • The current is up the page • The force is to the left**Force on a Wire**• B is into the page • The current is down the page • The force is to the right**Force on a Wire, equation**• The magnetic force is exerted on each moving charge in the wire • The total force is the sum of all the magnetic forces on all the individual charges producing the current FM = B I l sin q • q is the angle between and the direction of I • The direction is found by the right hand rule, placing your fingers in the direction of I instead of**Example 5**• A flat, horizontal rectangular loop of wire is positioned, as shown in Fig. 19.32a, in a 0.10-T uniform vertical magnetic field. The sides of the rectangle are F-C equal to 30 cm and C- equal to 20 cm. Determine the total force acting on the loop when it carries a current of 1.0 A. • Given: B = 0.10 T, F-C = 30 cm, C-D = 20 cm, and I = 1.0 A • Find: FM.**Example 5**• Solution: We've got a current-carrying loop in a S-field – FM = IlB sin q • First, make a drawing • Current travels from the positive terminal of the battery clockwise around the circuit. The directions of the forces on each segment are arrived at via vxB and are indicated in the diagram. Because the forces on segments F-C and D-E are equal and opposite, they cancel. The total force acting on the loop is the force acting on segment C-D**Example 5**FM= IlB sin q = (1.0 A) (0.20m) (0.10 T) (sin 90°) FM = 0.020 N**f**f Torque on a Current Loop • Applies to any shape loop • N is the number of turns in the coil • Torque has a maximum value of NIAB • When f = 90° • Torque is zero when the field is parallel to the plane of the loop**Electric Motor**• An electric motor converts electrical energy to mechanical energy • The mechanical energy is in the form of rotational kinetic energy • An electric motor consists of a rigid current-carrying loop that rotates when placed in a magnetic field**Electric Motor, 2**• The torque acting on the loop will tend to rotate the loop to smaller values of q until the torque becomes 0 at q = 0° • If the loop turns past this point and the current remains in the same direction, the torque reverses and turns the loop in the opposite direction**Electric Motor, 3**• To provide continuous rotation in one direction, the current in the loop must periodically reverse • In ac motors, this reversal naturally occurs • In dc motors, a split-ring commutatorand brushes are used • Actual motors would contain many current loops and commutators**Electric Motor, final**• Just as the loop becomes perpendicular to the magnetic field and the torque becomes zero, inertia carries the loop forward and the brushes cross the gaps in the ring, causing the current loop to reverse its direction • This provides more torque to continue the rotation • The process repeats itself ***This is obviously much easier if the CURRENT itself periodically reverses – AC electricity instead of DC*****Electromagnetic Induction**• Moving charge(s) creates (induces) a magnetic field (B) • A moving (changing) magnetic field (B) creates (induces) a current (I) • Whether B is increasing or decreasing near a wire determines the current’s direction.**Michael Faraday**• 1791 – 1867 • Great experimental scientist • Invented electric motor, generator and transformers • Discovered electromagnetic induction • Discovered laws of electrolysis

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