MAGNETIC FIELD AND ELEKTROMAGNETIC INDUCTION

MAGNETIC FIELD AND ELEKTROMAGNETIC INDUCTION
S N PHYSICS SMK PERGURUAN CIKINI

Isi dengan Judul Halaman Terkait
MAGNETIC FIELD Magnetic field is usually represented by the imaginary lines called magnetic field lines or magnetic force lines. These magnetic field lines have direction that come out from the north pole and entering south pole of magnet as shown in the following figure. S N S N Hal.: 2 Isi dengan Judul Halaman Terkait

MAGNETIC FIELD S N Press here, Please! This figure shows how the magnetic field of a bar magnet can be traced with the aid of a compass. Hal.: 3 Isi dengan Judul Halaman Terkait

MAGNETIC FIELD There are three rules of magnetic field lines, these are : a. The magnetic field lines never cut (cross) each other. b. The magnetic field lines always come out from the north pole and entering the south pole and form closed curves. c. If the magnetic field lines at a certain location are dense, then the magnetic field at that location is strong, vice versa if the magnetic field lines at a certain location are wide apart, then the magnetic field at that location is weak. Hal.: 4 Isi dengan Judul Halaman Terkait

MAGNETIC INDUCTION Basically, the source of magnetic field are not only shaped of permanent magnet, but can also in the form of electromagnet, that is, the magnet produced by electric current or moving electric charge. S U S U S U The result of Oersted’s experiment Hal.: 5 Isi dengan Judul Halaman Terkait

BIOT-SAVART’S LAW Note: B = magnetic induction (T) mo = vacum permeability (4p x 107 Wb/Am) I = electric current (A) r = radius of the circular path (m) P Hal.: 6 Isi dengan Judul Halaman Terkait

BIOT-SAVART’S LAW The magnetic induction at point O can be determined by the following equation: O r If there are N windings of the circular wire, then that equa-tion become. Note: N = the widing number r = the radius of wire ( m ) Hal.: 7 Isi dengan Judul Halaman Terkait

BIOT-SAVART’S LAW Meanwhile, the magnetic induction at point S as follows: S O r a q P Note: a = the distance of point p with point s ( m ) r = the radius of wire ( m ) q = the angle of SP with SO Hal.: 8 Isi dengan Judul Halaman Terkait

MAGNETIC INDUCTION IN SOLENOID
Source: The magnetic induction in the middle of solenoid can be determined by the following equation: The magnetic induction at that out of both ends as follows. Note: i = electric current ( A ) l = length of solenoid ( m ) N = winding number Hal.: 9 Isi dengan Judul Halaman Terkait

Magnetic induction on the toroid
The magnetic induction on the toroid can be can be determined by the equation as follows: r B Note: r = radius of toroid ( m ) I = electric current ( A ) N = winding number Source: Hal.: 10 Isi dengan Judul Halaman Terkait

THE MAGNETIC INDUCTION
Example What is the magnetic induction at distance of 5 cm from the center of straight wire carrying 3 A current? Solution mo = 4 p x 107 Tm/A I = 3 A r = 5 cm = 0.05 m B = ….? Thus, the magnetic induction produced is 1.2 x 105 T. Hal.: 11 Isi dengan Judul Halaman Terkait

LORENTZ FORCE Lorentz force in the straight wire carrying electric current If a wire with length of l flown by electric current I exists in a magnetic field B, then the wire will undergo Lorentz force or magnetic force which its direction can be determined by the right-hand rule. S N Hal.: 12 Isi dengan Judul Halaman Terkait

LORENTZ FORCE Source : Note: FL= lorentz force (N) B = magnetic induction (T)  = the angel between B and I I = electric current (A) l = length of wire (m) The thumb direction represents the direction of electric current, the fingers direction represents the directions of magnetic induction, and the palm side direction represents the direction of Lorentz force. Hal.: 13 Isi dengan Judul Halaman Terkait

LORENTZ FORCE Lorentz force in two parallel wires carrying electric current r X F1 F2 B2 B1 I1 I2 Note: r = distance of both wires (m) I = electric current (A) l = length of wire (m) Hal.: 14 Isi dengan Judul Halaman Terkait

LORENTZ FORCE Lorentz force on moving electric charges If an electric charge is moving in the magnetic field, then it will undergo Lorentz force which its magnitude can be determined by the following equation: + X X X v FL B - Positive charge Negative charge Note : B = magnetic induction (T)  = the angel between B and v q = electric charges (C) v = particle’s speed (m/s) Hal.: 15 Isi dengan Judul Halaman Terkait

LORENTZ FORCE If the direction of v is parallel to the direction of magnetic induction B, then the Lorentz force on the charged particle is zero, so that it moves linearly, yet if the direction of v is perpendicular to B, then the Lorentz force on the charged particle is FL=Bqv and follows a circular path with radius R. Thus, the magnitude of Lorentz force FL is equal to the centripetal force Fs. Therefore, Note: R = radius of path (m) m = mass of particle (kg) q = anguler speed of particle (rad/s) Hal.: 16 Isi dengan Judul Halaman Terkait

Moment of lorentz force Isi dengan Judul Halaman Terkait
When a conductor wire shaped of the coil with cross sectional area of A is flown by electric current in magnetic field, then the coil will undergo moment of lorentz force. Note:  = moment of force (Nm) I = electric current in the coil (A) B = magnetic induction (T) A = area of coil (m2)  = the angle between B with the coil plane Hal.: 17 Isi dengan Judul Halaman Terkait

LORENTZ FORCE Example A wire with 2 meters in length is flown by electric current 50 A. if the wire undergoes the magnetic force of 1.5 N in the magnetic field which is homogeneous with B = 0.03 T, then determine the angle between B and I? Solution FL = 1.5 N B = 0.03 T I = 50 A l = 2 m a = ….? Thus, the angle between B and I is 30o. Hal.: 18 Isi dengan Judul Halaman Terkait

THE MAGNETISM CHARACTERISTICS OF MATERIAL
Based on how the materials react to a magnetic field, then magnetic materials can be distinguished into diamagnetic materials, paramagnetic materials, and ferromagnetic materials. Diamagnetic materials are those that are replled slightly by magnetic field, the example are gold, cooper, etc. Paramagnetic materials are those that attracted with very low force in the magnetic field, the example are alumunium, magnesium, etc. Ferromagnetic materials are those that attracted strongly in magnetic field. Hal.: 19 Isi dengan Judul Halaman Terkait

ELECTROMAGNETIC INDUCTION Magnetic flux a N B A Note:  = magnetic flux (Wb) B = magnetic induction (T) A = surface area (m2)  = the angle between B with the plane’s normal line Hal.: 20 Isi dengan Judul Halaman Terkait

ELECTROMAGNETIC INDUCTION Faraday-Lenz’s law Source: Note: eind = induction electromotive force (volt) D = the magnetic flux changing (Wb) N = winding number Dt = the time interval (s) Hal.: 21 Isi dengan Judul Halaman Terkait

Electromagnetic induction
Example A coil has 100 windings and in 0.01 s emerges a magnetic flux change of 10-4 Wb, calculate the induction electromotive force at the coil ends? Solution N = 100 df = 10-4 Wb dt = 0.01 s eind = …..? Thus, the induction electromotive force at the coil ends is 1 volt. Hal.: 22 Isi dengan Judul Halaman Terkait

ELECTRIC GENERATOR Source: Note: N = winding number B = magnetic induction (T) A = area of coil plane (m2) = anguler velocity (rad/s) t = time (s) Hal.: 23 Isi dengan Judul Halaman Terkait

ELECTRIC GENERATOR The scheme of AC generator 1 1. Slip ring 2. Loop 3. External circuit 4. Brushes 5. Rotor luar 2 3 4 5 Source: Hal.: 24 Isi dengan Judul Halaman Terkait

ELECTRIC GENERATOR The scheme of DC generator 1 1. Sikat 2. Armature 3. Comutator 2 3 Hal.: 25 Isi dengan Judul Halaman Terkait

INDUCTANCE The value of self-induction electromagnetive force occurs in the circuit or coil depends on the rate change of the current. Note: eind = self-induction electromotive force (volt) DI = the electric current changing (A) L = inductance (H) Dt = the time interval (s) Hal.: 26 Isi dengan Judul Halaman Terkait

INDUCTANCE Example A coil has inductance of 5 H and a resistor has resistance of 20 W. Both are set in the voltage source of 100 volt. Calculate the energy stored on the coil if the current reach maximum value? Solution e = 100 volt R = 20 W L = 5 H W = ….? Thus, the energy stored in the coil is 63 J Hal.: 27 Isi dengan Judul Halaman Terkait

TRANSFORMER Note: Vp = primary voltage (volt) Vs = secondary voltage (volt) Np = primary winding number Ns = secondary winding number Ip = primary electric current (A) Is = secondary electric current (A) Hal.: 28 Isi dengan Judul Halaman Terkait

TRANSFORMER Transformer efficiency Note: = transformer efficiency P1 = primary power (watt) P2 = secondary power (watt) Hal.: 29 Isi dengan Judul Halaman Terkait

EXERCISE One of a magnetic pole is moved entering a coil. The direction of induction current emerges in the coil is anti-clockwise. a. which is the pole entered? b. what is the direction of induction current if the magnetic is pulled out? 2. Explain the working principle of generators and what are the differences between alternating and direct current generators? 8 W 4 H 24 volt S 3. In such a circuit, determine the time constant of the circuit and energy stored in the inductor when the current reaches its maximum value? Hal.: 30 Isi dengan Judul Halaman Terkait

thank you Hal.: 31 Isi dengan Judul Halaman Terkait

MAGNETIC FIELD AND ELEKTROMAGNETIC INDUCTION

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