Physics Magnetism - SHS STEM 2nd semester

GENERAL PHYSICS: MAGNETISM INTRODUCTION | MAGNETIC FIELD, FORCE, FLUX | FARADAY’S LAW | LENZ’S LAW | BIOT-SAVART LAW | AMPERE’S LAW | INDUCED ELECTRIC FIELD | AC & LC CIRCUITS SHS | STEM | SECOND SEMESTER | 2024 MAGNETISM MAGNETIC FIELD...

GENERAL PHYSICS: MAGNETISM INTRODUCTION | MAGNETIC FIELD, FORCE, FLUX | FARADAY’S LAW | LENZ’S LAW | BIOT-SAVART LAW | AMPERE’S LAW | INDUCED ELECTRIC FIELD | AC & LC CIRCUITS SHS | STEM | SECOND SEMESTER | 2024 MAGNETISM MAGNETIC FIELD MAGNETIC POLES  The space around the magnet or any other  Parts of the magnets where the magnetic field magnetic object where magnetic field/force is strongest. acts.  Moving farther away from the poles MAGNETIC FIELD LINES results in the weakening of the  A series of lines that represent magnetic field. magnetic field/force.  Continuous loops  North and South poles  Outside the magnet – point out of the north  Magnetic monopoles do not exist; always in pole and then they curve toward the south dipoles. pole (N to S).  North and south pole can never separate.  Inside the magnet – continue on straight  Unlike poles attract, like poles repel. toward the north pole again (S to N).  Three dimensional, curving around the MAGNETIZATION & DEMAGNETIZATION magnet in all directions around the length of  The magnetic properties of a magnet usually the bar. wear out with the passage of time or through For the poles  North – leaving ; South – various demagnetizing techniques. entering  Magnets can be demagnetized by For the  Closer together, hammering, heating, and being exposed to an spacing magnetic field is strong alternating current.  Far from one another,  These techniques involve mixing up the magnetic field is weaker arrangement of molecules to cancel the polarity of the entire material. However, For the  Closer to the source, molecules can be rearranged again through distance magnetic field is strong magnetization, thus bringing back the  Further from the source, magnetic field is weaker material’s overall polarity.  A demagnetized magnet can be magnetized again by exposing it to another magnet with a strong magnetic field.  Types of magnets: permanent, natural, temporary, electromagnets.  Methods of demagnetization: hammering, burning/heating, and being exposed to alternating current.  Solenoid – loop of wire or coil OERSTED DISCOVERY Magnetic Field Formula  Hans Cristian Oersted 𝑀𝑎𝑔 𝑓𝑖𝑒𝑙𝑑 𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒  This discovery led to the concept of (𝑝𝑒𝑟𝑚𝑒𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑓𝑟𝑒𝑒 𝑠𝑝𝑎𝑐𝑒)(𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒) electromagnetism. = 2𝜋(𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒) ELECTROMAGNETISM 𝝁𝟎 𝑰  Moving electric charges (current) induces a 𝑩= 𝟐𝝅𝒓 magnetic field perpendicular to the flow of current.  𝐵 – magnetic field magnitude (Tesla, T)  When current is turned off, no MAGNETIC  𝜇0 – permeability of free space FIELD EXISTS, and no force exists. (4𝜋𝑥10−7 𝑇 ∙ 𝑚/𝐴)  Current = Magnetic Field = Magnetic Force  𝐼 – magnitude of the electric current  Electromagnet – conductor of current (Amperes, A)  𝑟 – distance (m) Note: 1 Tesla = 1𝑥104 𝐺𝑎𝑢𝑠𝑠 (𝐺) 𝑜𝑟 10,000 𝐺 fau GENERAL PHYSICS: MAGNETISM INTRODUCTION | MAGNETIC FIELD, FORCE, FLUX | FARADAY’S LAW | LENZ’S LAW | BIOT-SAVART LAW | AMPERE’S LAW | INDUCED ELECTRIC FIELD | AC & LC CIRCUITS SHS | STEM | SECOND SEMESTER | 2024 RIGHT HAND GRIP RULE MAGNETIC FORCE  Used to determine the direction of B and I.  The magnetic force is a consequence of the  Thumb – denotes the direction of the electromagnetic force, one of the four current. fundamental forces of nature, and is caused  4 fingers – denotes the direction of the by the motion of charges. magnetic field.  Magnetic field results to magnetic force; no magnetic field, no magnetic force.  Two objects containing charge with the same direction of motion have a magnetic attraction force between them. Similarly, objects with charge moving in opposite directions have a repulsive force between them.  Factors to consider in determining the magnetic force: The magnitude of the magnetic force between them depends on how much charge (𝒒) is in how much  Current coming  Current going motion (𝒗) in each of the two objects and out of the page into the page how far apart (𝑩) they are. towards you – you away from you – can see the arrow you can see the Magnetic Force Formula tip/ cross of the arrow’s tail-  A charge particle moving in magnetic field feathers/ 𝑭𝑴 = 𝒒𝒗𝑩 ; 𝑭𝑴 = 𝒒𝒗𝑩 𝒔𝒊𝒏𝜽  𝑞 – charge (C)  𝑣 – velocity of the charge (m/s)  𝐵 – strength of magnetic field (Tesla, 𝑇 = 𝑁/𝐴 ∙ 𝑚)  The magnetic force is proportional to q and to the magnitude of the vector cross product q x B. EXAMPLE 1 What is the magnitude of the magnetic field 100 𝑚 away from a wire carrying a 6.00 𝐴 current? If the RIGHT HAND RULE (RHR) current has a vector direction out of the page (or  Used to determine the direction of the screen), what is the direction of the magnetic field? magnetic force.  Thumb – direction of the force (𝑭𝑴 )  Index finger – direction of the moving 𝑟 = 100 𝑚 charges or current (𝒗) Given 𝐼 = 6.00 𝐴, 𝑜𝑢𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑎𝑔𝑒  Middle finger – direction of the magnetic 𝜇𝑜 = 4𝜋 𝑥 10−7 𝑇𝑚/𝐴 field (𝑩) Asked 𝐵=? 𝜇0 𝐼 FM Formula 𝐵= 2𝜋𝑟 𝜇0 𝐼 𝐵= v 2𝜋𝑟 Solution (4𝜋 𝑥 10−7)(6.00 𝐴) 𝐵= 2𝜋(100 𝑚) 𝑩 = 𝟏. 𝟐 𝒙 𝟏𝟎−𝟖 𝑻 B Final Answer Counterclockwise (Right Hand Grip Rule) fau GENERAL PHYSICS: MAGNETISM INTRODUCTION | MAGNETIC FIELD, FORCE, FLUX | FARADAY’S LAW | LENZ’S LAW | BIOT-SAVART LAW | AMPERE’S LAW | INDUCED ELECTRIC FIELD | AC & LC CIRCUITS SHS | STEM | SECOND SEMESTER | 2024 EXAMPLE 1 MOVEMENT OF PARTICLES IN A MAG FIELD A 6.4 𝑥 10−11 𝐶 charged proton with a velocity of  The movement of the charge when placed in 9.7 𝑥 104 𝑚/𝑠 is moving through a space with a a magnetic field is in a circular motion. magnetic field strength of 1.2 𝑇. From the magnetic  A magnetic field is a region in which a moving field pointing to the east, how much force would the charge experiences a force. Since electrons force feel? are moving charges, they will experience a force when they enter a magnetic field.  If a charged particle enters a magnetic field 𝑞 = 6.4 𝑥 10−11 𝐶 perpendicular to the field, it will Experience a Given 𝑣 = 9.7 𝑥 104 𝑚/𝑠, 𝑠𝑜𝑢𝑡ℎ force at right angles to its motion and it will 𝐵 = 1.2 𝑇, 𝑒𝑎𝑠𝑡 be steered into a circular path. Asked 𝐹𝑀 , 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛  Since this is a circular path, there will be Formula 𝐹𝑀 = 𝑞𝑣𝐵 𝑠𝑖𝑛𝜃 radius. 𝐹𝑀 = 𝑞𝑣𝐵 𝑠𝑖𝑛𝜃 Solution 𝐹𝑀 = 𝒎𝒗 𝑹= (6.4 𝑥 10 −11 )( 9.7 𝑥 10 4 )( 1.2)𝑠𝑖𝑛90 𝒒𝑩 −𝟔 𝑭𝑴 = 𝟕. 𝟒𝟓 𝒙 𝟏𝟎 𝑵  The radius of the circular Final Answer Out of the page/Towards you path can be calculated by (Right Hand Rule) dividing the product of the mass and velocity of the EXAMPLES FOR RHR2 particle and the product of the absolute value of the 1. A positive charged particle is moving to the left and the magnetic field it is in is towards the outside. charge and the magnitude What would be the direction of the magnetic force? of the magnetic field.  Upward/North 2. A positive charged particle is moving up and the Cyclotron Formula magnetic field it is in is towards you. What would be the direction of the magnetic force?  The number of times or rotations that a  To the right/ East charged particle will make in the magnetic 3. Charge – Positive ; Force – North ; Magnetic Field field is given by the cyclotron formula. – into the page ; Velocity – ? 𝒒𝑩  To the right/East 𝒇= 4. Charge – Positive ; Force – North ; Magnetic Field 𝟐𝝅𝒎 – out the page ; Velocity – ?  𝑓 – Frequency, Hertz (Hz)  To the left/West  𝑞 – absolute value of the charge, 5. Charge – Positive ; Magnetic Field – into the page Coulombs (C) ; Velocity – South ; Force – ?  To the right/East  𝐵 – magnetic field strength, Tesla (T)  𝑚 – mass, mass of the particle (kg)  The standard unit is Hertz (Hz), after the For a negative charge German physicist Heinrich Hertz.  For a negative charge, you have to reverse the direction of the force. The q and v will stay the same. MAGNETIC FLUX  Total number of magnetic field lines passing through a given coil or area. 𝝓𝑩 = 𝑩𝑨𝒄𝒐𝒔𝜽  𝐵 – magnetic field, Tesla (T)  𝐴 – surface area, 𝑚2  𝜃 – angle between the area vector/normal to the surface and magnetic field.  Unit: 𝑇 ∙ 𝑚2 = 𝑊𝑒𝑏𝑒𝑟 (𝑊𝑏) fau GENERAL PHYSICS: MAGNETISM INTRODUCTION | MAGNETIC FIELD, FORCE, FLUX | FARADAY’S LAW | LENZ’S LAW | BIOT-SAVART LAW | AMPERE’S LAW | INDUCED ELECTRIC FIELD | AC & LC CIRCUITS SHS | STEM | SECOND SEMESTER | 2024 Case 1 𝐴 = 𝑙𝑤  If the surface or the coil is perpendicular to 𝐴 = 0.051 𝑚 (0.068 𝑚) the magnetic field lines: 𝐴 = 3.47𝑥10−3 𝑚2 Solution  𝜃 = 0° 𝑜𝑟 180°  𝜙𝐵 = 𝑚𝑎𝑥 𝜙𝐵 = 𝐵𝐴𝑐𝑜𝑠𝜃 𝜙𝐵 = (0.02)(3.47𝑥10−3 )𝑐𝑜𝑠47° Final Answer 𝝓𝑩 = 𝟒. 𝟕𝟑𝒙𝟏𝟎−𝟓 𝑾𝒃 Magnetic Force on a Current-carrying Wire  If we put a current carrying wire in a magnetic field, the wire as a whole will experience a Case 2 magnetic force.  If the surface or the coil is parallel to the magnetic field lines: 𝑭𝑩 = 𝑰𝑳𝒙𝑩 ; 𝑭𝑩 = 𝑰𝑳𝒙𝑩𝒔𝒊𝒏𝜽  𝜃 = 90°  𝐹𝐵 – magnetic force, N  𝜙𝐵 = 0  𝐼 – current, A  𝐿 – length of the wire, m  𝐵 – magnetic field, T  The direction can be determined by the Palm method or RHR2 PALM METHOD Case 3  Used to determine the direction of the  If the surface or the coil is at a certain angle magnetic force. from the magnetic field lines:  Palm – direction of the magnetic force  𝜃 = 𝑔𝑖𝑣𝑒𝑛 𝑎𝑛𝑔𝑙𝑒  Thumb – direction of the current  𝜙𝐵 = 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝑚𝑎𝑥𝑖𝑚𝑢𝑚  4 fingers – direction of the magnetic field EXAMPLE 1 A current of 20 A flows East through a 50 cm long EXAMPLE 1 wire. A magnetic field of 4.0 T is directed into the Dimension of a rectangular loop is given as 0.051 𝑚 page. What is the magnitude and direction of the and 0.068 𝑚. B and θ are 0.02𝑇 and 47° magnetic force acting on the wire? respectively. Calculate the magnetic flux through the given surface. 𝐼 = 20 𝐴 (𝑒𝑎𝑠𝑡) Given 𝐵 = 0.02 𝑇 (𝑖𝑛𝑡𝑜 𝑡ℎ𝑒 𝑝𝑎𝑔𝑒) 𝐵 = 0.02 𝑇 𝐿 = 50 𝑐𝑚 = 0.50 𝑚 𝜃 = 47° Asked 𝐹𝐵 = ? Given 𝑙 = 0.051 𝑚 Formula 𝐹𝐵 = 𝐼𝐿𝑥𝐵𝑠𝑖𝑛𝜃 𝑤 = 0.068 𝑚 𝐹𝐵 = 𝐼𝐿𝑥𝐵𝑠𝑖𝑛𝜃 Asked Solution 𝜙𝐵 = ? 𝐹𝐵 = (20)(0.50)(4.0)𝑠𝑖𝑛90° Formula 𝜙𝐵 = 𝐵𝐴𝑐𝑜𝑠𝜃 Final Answer 𝑭𝑩 = 𝟒𝟎 𝑵, 𝑵𝒐𝒓𝒕𝒉 fau GENERAL PHYSICS: MAGNETISM INTRODUCTION | MAGNETIC FIELD, FORCE, FLUX | FARADAY’S LAW | LENZ’S LAW | BIOT-SAVART LAW | AMPERE’S LAW | INDUCED ELECTRIC FIELD | AC & LC CIRCUITS SHS | STEM | SECOND SEMESTER | 2024 FARADAY’S LAW  Step up – the output voltage is higher than  Faraday’s Law of Electromotive Induction the input.  “The induced electromotive force in a closed  Step down – the output voltage is lower loop is equal to the negative times rate of than the input. change of magnetic flux inside the loop.” EXAMPLE 1 ∆𝝓𝑩 𝝓𝑩𝒇 − 𝝓𝑩𝒊 A flat, square coil of 0.25 𝑚 on each side with 𝑁 = 6 𝜀=− →𝜀=− loops is experiencing an induced emf due to the 𝚫𝒕 𝚫𝒕 magnetic field of 1.2 𝑇 passing through it. The plane  If the wire has multiple coils, the EMF is of the coil is perpendicular to the magnetic field: the multiplied by the number of coils (N): filed lines point out of the page. The magnetic field ∆𝝓𝑩 increases uniformly from 1.2 𝑇 to 1.5 𝑇 in 2𝑠. What 𝜀 = −𝑵 is the induced emf in the coil, while the change of 𝚫𝒕 magnetic field is taking place? FACTORS CHANGING THE INDUCED EMF ∆𝝓𝑩 𝑠 = 0.25 𝑚 𝑜𝑛 𝑒𝑎𝑐ℎ 𝑠𝑖𝑑𝑒 𝜀 = −𝑵 𝝓𝑩 = 𝑩𝑨𝒄𝒐𝒔𝜽 𝑁 = 6 𝑙𝑜𝑜𝑝𝑠 𝚫𝒕 𝐵𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 1.2 𝑇 Given 𝐵𝑓𝑖𝑛𝑎𝑙 = 1.5 𝑇 1. Change in B 4. Number of coils (N) 2. Change in A 5. Time 𝑡 = 2𝑠 3. Change in 𝜃 of B & A 𝜃 = 0° Asked 𝜀, 𝑒𝑚𝑓 = ? How do we change the Magnetic Flux? ∆𝜙𝐵 𝜀=−  In order to change the magnetic flux, there Δ𝑡 Formula 𝜙𝐵 = 𝐵𝐴𝑐𝑜𝑠𝜃 should be a change in the magnetic field.  To change the magnetic field, according to 𝐴 = 𝑠2 Faraday’s Law, there should be movement. For the area, 𝐴 = 𝑠2 𝐴 = (0.25 𝑚)2 𝐴 = 0.0625 𝑚2 For the initial magnetic flux, 𝜙𝐵 = 𝐵𝐴𝑐𝑜𝑠𝜃 𝜙𝐵 = (1.2 𝑇)(0.0625 𝑚2 )𝑐𝑜𝑠0° 𝜙𝐵 = 0.075 𝑊𝑏 Solution For the final magnetic flux,  Without movement, there won’t be induced 𝜙𝐵 = 𝐵𝐴𝑐𝑜𝑠𝜃 EMF/current. 𝜙𝐵 = (1.5 𝑇)(0.0625 𝑚2 )𝑐𝑜𝑠0° Using Faraday’s Law 𝜙𝐵 = 0.094 𝑊𝑏 To better understand Faraday’s law, consider its For the induced emf, applications on devices that form modern ∆𝜙𝐵 technology. 𝜀=− Δ𝑡 (0.094 𝑊𝑏 − 0.075 𝑊𝑏) 𝜀 = −6 2s 𝜺 = −𝟎. 𝟎𝟓𝟕 𝑽 Final Answer 𝜺 = −𝟎. 𝟎𝟓𝟕 𝑽 EXAMPLE 2 A circular coil of a diameter of 13 𝑐𝑚 and with 1498 turns is in a uniform magnetic field that is directed at an angle of 30° with respect to the plane of the coil. fau GENERAL PHYSICS: MAGNETISM INTRODUCTION | MAGNETIC FIELD, FORCE, FLUX | FARADAY’S LAW | LENZ’S LAW | BIOT-SAVART LAW | AMPERE’S LAW | INDUCED ELECTRIC FIELD | AC & LC CIRCUITS SHS | STEM | SECOND SEMESTER | 2024 If the magnetic field changes at a rate of 5 𝑇𝑒𝑠𝑙𝑎 per 𝜙𝐵 = 𝐵𝐴𝑐𝑜𝑠𝜃 1.5 𝑠𝑒𝑐𝑜𝑛𝑑, what will be the voltage of induced 𝐴 = 𝜋𝑟 2 electromotive force? The change in the magnetic flux is computed as follows: ∆𝜙 = 𝐵𝑓𝑖𝑛𝑎𝑙 𝐴 − 𝐵𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝐴 𝑑 = 13 𝑐𝑚 = 0.13 𝑚 Because the areas are the same, 𝑁 = 1498 the expression can be simplified as 𝜃 = 30° follows: 𝑤𝑖𝑡ℎ 𝑟𝑒𝑠𝑝𝑒𝑐𝑡 𝑡𝑜 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑖𝑙 Given ∆𝜙 = (𝐵𝑓𝑖𝑛𝑎𝑙 − 𝐵𝑖𝑛𝑖𝑡𝑖𝑎𝑙 )(𝐴) 𝜃 = 60° 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝐵 𝑎𝑛𝑑 𝐴 Solution ∆𝜙 𝐵 =5𝑇 = (0.35 𝑇 − 0.10 𝑇)(𝜋)(0.03 𝑚)2 𝑡 = 1.5 𝑠 ∆𝜙 = 7.1𝑥10−4 𝑊𝑏 Asked 𝜀, 𝑒𝑚𝑓 = ? ∆𝜙𝐵 The average EMF is computed as 𝜀 = −𝑁 Δ𝑡 follows: 𝜙𝐵 = 𝐵𝐴𝑐𝑜𝑠𝜃 ∆𝜙𝐵 Formula 𝜀=𝑁 𝐴 = 𝜋𝑟 2 Δ𝑡 𝑑 7.1𝑥10−4 𝑊𝑏 𝑟= 𝜀 = (50) 2 2x10−3 𝑠 For diameter to radius, Final Answer 𝜺 = 𝟏𝟕. 𝟕 𝑽 𝑑 0.13 𝑚 𝑟= = = 0.065 𝑚 2 2 LENZ’S LAW  Heinrich Lenz For the area of the coil,  “The direction of the induced current is such 𝐴 = 𝜋𝑟 2 that its own magnetic field opposes the 𝐴 = 𝜋(0.065 𝑚)2 change that induced it.” 𝐴 = 0.013 𝑚2  Faraday’s Law: Solution 𝑑𝜙𝐵 For the emf, 𝜀 = −𝑁 𝑑𝑡 𝑁𝑑𝜙𝐵  The minus sign in the Faraday’s law of 𝜀=− 𝑑𝑡 induction is due to the fact that the induced 𝑁 𝜀=− 𝑑(𝐵𝐴𝑐𝑜𝑠𝜃) emf will always oppose the change. 𝑡  The current from the induced emf will 𝑑𝐵 𝜀 = −𝑁𝐴𝑐𝑜𝑠𝜃 produce a magnetic field, which will 𝑑𝑡 5𝑇 always oppose the original change in the = −(1498)(0.013 𝑚2 )𝑐𝑜𝑠60° magnetic flux. 1.5 𝑠 Final Answer 𝜺 = −𝟑𝟐. 𝟒𝟔 𝑽 EXAMPLE 3 A circular coil with a radius of 3 𝑐𝑚 is made up of 50 loops. Its orientation makes the magnetic field lines normal to the area of the coil. If the magnetic field is varied from 0.10 𝑇 to 0.35 𝑇 in 2 𝑚𝑖𝑙𝑙𝑖𝑠𝑒𝑐𝑜𝑛𝑑𝑠, what is the average induced EMF in the coil?  Lenz’s law serves as a guide in figuring out which way the current in a loop goes. 𝑟 = 3 𝑐𝑚 = 0.03 𝑚  It helps in remembering that the induced emf 𝑁 = 50 has an induced current, so there is also an Given 𝐵𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 0.10 𝑇 induced magnetic field. 𝐵𝑓𝑖𝑛𝑎𝑙 = 0.35 𝑇  An induced magnetic field does not create nor 𝑡 = 2 𝑚𝑠 = 2𝑥10−3 𝑠 add energy to a system. Asked 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝜀 = ?  When the magnetic field is already increasing, ∆𝜙𝐵 the induced magnetic field prevents energy Formula 𝜀 = −𝑁 from being added on and on. Δ𝑡 fau GENERAL PHYSICS: MAGNETISM INTRODUCTION | MAGNETIC FIELD, FORCE, FLUX | FARADAY’S LAW | LENZ’S LAW | BIOT-SAVART LAW | AMPERE’S LAW | INDUCED ELECTRIC FIELD | AC & LC CIRCUITS SHS | STEM | SECOND SEMESTER | 2024 CASE 3: 1. Moving a magnet near a coil will induce a Move South Pole Closer to Coil current. 𝐼 → 𝐵𝑖𝑛𝑑 2. Induced magnetic field always tries to keep the flux in the loop constant. CASE 1: Move North Pole Closer to Coil  Since the pole is South, the original magnetic field (𝐵) is entering the South pole/ to the right.  If the magnet is moved closer to the coil, there would be an increase in the magnetic field.  That increase (∆𝐵) would be in the same  Since the pole is North, the original magnetic direction as the original (𝐵). field (𝐵) is leaving the North pole/away/to the  The induced magnetic field (𝐵𝑖𝑛𝑑 ) will then left. oppose the direction of the increase (∆𝐵).  If the magnet is moved closer to the coil,  The current is flowing upward. there would be an increase in the magnetic field. CASE 4:  That increase (∆𝐵) would be in the same Move South Pole Away from Coil direction as the original (𝐵).  The induced magnetic field (𝐵𝑖𝑛𝑑 ) will then oppose the direction of the increase (∆𝐵).  The current is flowing downward. CASE 2: Move North Pole Away from Coil  Since the pole is South, the original magnetic field (𝐵) is entering the South pole/ to the right.  If the magnet is moved away to the coil, there would be a decrease in the magnetic field.  That decrease (∆𝐵) would be in the opposite direction as the original (𝐵).  Since the pole is North, the original magnetic  The induced magnetic field (𝐵𝑖𝑛𝑑 ) will then field (𝐵) is leaving the North pole/away/to the oppose the direction of the decrease (∆𝐵). left.  The current is flowing downward.  If the magnet is moved away to the coil, there would be a decrease in the magnetic field. BIOT-SAVART LAW  That decrease (∆𝐵) would be in the opposite  Jean-Baptiste Biot and Felix Savart direction as the original (𝐵).  Used to determine the magnetic field at a  The induced magnetic field (𝐵𝑖𝑛𝑑 ) will then point due to a current. oppose the direction of the decrease (∆𝐵).  This law predicts the magnetic field at a point  The current is flowing upward. P, a distance r, away from the current I in a wire.  This is also applied in any problems with asymmetrical elements. Steps: 1. Calculate the magnetic field generated by a current element fau GENERAL PHYSICS: MAGNETISM INTRODUCTION | MAGNETIC FIELD, FORCE, FLUX | FARADAY’S LAW | LENZ’S LAW | BIOT-SAVART LAW | AMPERE’S LAW | INDUCED ELECTRIC FIELD | AC & LC CIRCUITS SHS | STEM | SECOND SEMESTER | 2024 2. Sum up the magnetic fields generated by a For an infinitely long, straight wire carrying a multitude of current elements at the same current I, the magnetic field at a point P and a point. distance a away from the wire: 𝝁𝒐 𝑰𝒅𝒍𝒔𝒊𝒏𝜽 𝝁𝒐 𝑰 𝒅𝑩 = 𝒅𝑩 = 𝟒𝝅 𝒓𝟐 𝟐𝝅𝒂  𝜇𝑜 – permeability (4𝜋𝑥10−7 𝑇𝑚/𝐴)  𝜇𝑜 – permeability (4𝜋𝑥10−7 𝑇𝑚/𝐴)  𝑑𝑙 – length of the wire segment, m  𝐼 – current, A  𝐼 – current, A  𝑎 – distance, m  𝑟 – distance, m  𝜃 – angle between the wire segment to a EXAMPLE 2 point P A long straight wire has a current of 200 𝐴. At what distance 𝒂 from the wire is the magnitude of the EXAMPLE 1 magnetic field equal to 0.4 𝑇? A long straight wire has a current of 4 𝐴. Calculate the magnetic field produced by the current from a 𝐵 = 0.4 𝑇 1 − 𝑐𝑚 wire at points A and B as given by the Given 𝐼 = 200 𝐴 distances in the figure. 𝜇𝑜 = 4𝜋 𝑥 10−7 𝑇𝑚/𝐴 Asked 𝑎=? 𝜇 𝐼 Formula 𝑑𝐵 = 𝑜 2𝜋𝑎 𝜇𝑜 𝐼 𝜇 𝐼 𝑑𝐵 = →𝑎= 𝑜 2𝜋𝑎 2𝜋𝐵 Solution −7 (4𝜋 𝑥 10 ) (200) 𝑎= 2𝜋(0.4 𝑇) Final Answer 𝒂 = 𝟏. 𝟎𝟎𝒙𝟏𝟎−𝟒 𝒎 𝐼 = 4𝐴 𝑑𝑙 = 1 𝑐𝑚 = 0.01 𝑚 Given 𝑟𝐴 = 10 𝑐𝑚 = 0.1 𝑚 The Magnetic Field at the Center of a Circular 𝑟𝐴−𝐵 = 20 𝑐𝑚 = 0.2 𝑚 Current Loop 𝜇𝑜 = 4𝜋𝑥10−7 𝑇𝑚/𝐴 Asked 𝑑𝐵 𝑎𝑡 𝑝𝑜𝑖𝑛𝑡 𝐴, 𝐵 = ? 𝝁𝒐 𝑵𝑰 𝒅𝑩 = 𝜇𝑜 𝐼𝑑𝑙𝑠𝑖𝑛𝜃 𝟐𝒂 Formula 𝑑𝐵 = 4𝜋 𝑟 2 At point A,  𝜇𝑜 – permeability (4𝜋𝑥10−7 𝑇𝑚/𝐴) −7 4𝜋𝑥10 (4)(0.01)𝑠𝑖𝑛90°  𝑁 – number of turns 𝑑𝐵 = 4𝜋 (0.1 𝑚)2  𝐼 – current, A 𝒅𝑩 = 𝟒𝒙𝟏𝟎−𝟕 𝑻  𝑎 – radius of the circle, m At point B, EXAMPLE 3 𝑟= √(0.1)2 + (0.2)2 A wire is shaped into a coil of 300 turns, each with Solution 𝑟 = 0.22 𝑚 a radius of 2.5 cm. If the current in the wire is 6 mA, 0.1 find the magnitude of the magnetic field at the 𝜃 = tan−1 center of the coil. 0.2 𝜃 = 26.57° 𝑅 = 2.5 𝑐𝑚 = 0.025 𝑚 −7 𝑁 = 300 𝑡𝑢𝑟𝑛𝑠 4𝜋𝑥10 (4)(0.01)𝑠𝑖𝑛26.57° Given 𝑑𝐵 = 4𝜋 (0.22 𝑚)2 𝐼 = 6 𝑚𝐴 = 6𝑥10−3 𝐴 𝒅𝑩 = 𝟑. 𝟕𝒙𝟏𝟎−𝟖 𝑻 𝜇𝑜 = 4𝜋 𝑥 10−7 𝑇𝑚/𝐴 𝒅𝑩 𝒂𝒕 𝑨 = 𝟒𝒙𝟏𝟎−𝟕 𝑻 Asked 𝑑𝐵 = ? Final Answer 𝜇 𝑁𝐼 𝒅𝑩 𝒂𝒕 𝑩 = 𝟑. 𝟕𝒙𝟏𝟎−𝟖 𝑻 Formula 𝑑𝐵 = 𝑜 2𝑎 fau GENERAL PHYSICS: MAGNETISM INTRODUCTION | MAGNETIC FIELD, FORCE, FLUX | FARADAY’S LAW | LENZ’S LAW | BIOT-SAVART LAW | AMPERE’S LAW | INDUCED ELECTRIC FIELD | AC & LC CIRCUITS SHS | STEM | SECOND SEMESTER | 2024 𝜇 𝑁𝐼 The Magnetic Field at the Center of a Very 𝑑𝐵 = 𝑜 2𝑎 Long Solenoid Solution −7 −3 (4𝜋𝑥10 ) (300) (6𝑥10 ) 𝑑𝐵 = 2(0.025 𝑚) 𝝁𝒐 𝑵𝑰 Final Answer 𝒅𝑩 = 𝟒. 𝟓𝟐𝒙𝟏𝟎−𝟓 𝑻 𝑩= = 𝝁𝒐 𝒏𝒍 𝑳 AMPERE’S LAW  𝜇𝑜 – permeability (4𝜋𝑥10−7 𝑇𝑚/𝐴)  Andre-Marie Ampere (1826)  𝑁/𝐿 = 𝑛 – number of turns per length  “For any closed loop, the dot product of the  𝑁 – number of turns magnetic field and the total distance (length  𝐿 – length of the wire elements) around the loop is equal to the  𝐼 – current, A product of the permeability constant and current enclosed by the loop.” EXAMPLE 2  This is applied in any problems with high A 1 − 𝑚𝑒𝑡𝑒𝑟 wire is shaped into a coil of 300 turns. symmetry. If the current in the wire is 6 𝑚𝐴, find the magnitude of the magnetic field at the center of the coil. ∮ 𝑩 ∙ 𝒅𝒍 = 𝝁𝒐 𝑰𝒆𝒏𝒄𝒍𝒐𝒔𝒆𝒅 𝐿 =1𝑚 Straight Current-carrying Wire with a 𝑁 = 300 𝑡𝑢𝑟𝑛𝑠 Given Circular Loop 𝐼 = 6 𝑚𝐴 = 6𝑥10−3 𝐴  When we SUM all of the current carrying 𝜇𝑜 = 4𝜋 𝑥 10−7 𝑇𝑚/𝐴 elements around the PATH of a circle, we get Asked 𝐵=? the circle’s circumference. 𝜇 𝑁𝐼 Formula 𝐵= 𝑜 𝐿 𝑩 ∮ 𝒅𝒍 = 𝝁𝒐 𝑰𝒆𝒏𝒄 𝑩(𝟐𝝅𝒓) = 𝝁𝒐 𝑰 𝜇𝑜 𝑁𝐼 𝐵= 𝐿  For straight current carrying wire with a Solution (4𝜋 𝑥 10−7 )(300)(6𝑥10−3 ) circular loop, the magnetic field generated by 𝐵= (1 𝑚) any current element or wire segment is the Final Answer 𝑩 = 𝟐. 𝟐𝟖𝒙𝟏𝟎−𝟔 𝑻 same. SUMMARY  Once again, we see we get the equation for Biot-Savart Law Ampere’s Law the magnetic field around a long straight wire.  The Biot-Savart  Ampère's Law is 𝝁𝒐 𝑰 𝑩= Law is used to used to find the 𝟐𝝅𝒓 calculate the magnetic field magnetic field (B) around a closed EXAMPLE 1 produced by a loop due to the Compute for the magnetic field along a straight wire steady current current passing that has a circular loop around it with a radius of element through the loop. 0.05 𝑚. When an ammeter was used, 2𝐴 appeared (infinitesimally a reading within the closed loop. small segment of current-carrying 𝑟 = 0.05 𝑚 wire). Given 𝐼 =2𝐴  The Biot-Savart  Ampère's Law is 𝜇𝑜 = 4𝜋 𝑥 10−7 𝑇𝑚/𝐴 Law can be particularly Asked 𝐵=? applied to any powerful for 𝜇𝑜 𝐼 Formula 𝐵= current solving problems 2𝜋𝑟 distribution, even with high 𝜇𝑜 𝐼 𝐵= when the symmetry, where 2𝜋𝑟 Solution symmetry is not the magnetic (4𝜋 𝑥 10−7 )(2 𝐴) 𝐵= present field has 2𝜋(0.05 𝑚) (asymmetrical). cylindrical or Final Answer 𝑩 = 𝟖𝒙𝟏𝟎−𝟔 𝑻 axial symmetry. fau GENERAL PHYSICS: MAGNETISM INTRODUCTION | MAGNETIC FIELD, FORCE, FLUX | FARADAY’S LAW | LENZ’S LAW | BIOT-SAVART LAW | AMPERE’S LAW | INDUCED ELECTRIC FIELD | AC & LC CIRCUITS SHS | STEM | SECOND SEMESTER | 2024  Derived from Biot-Savart: Ampère's Law is AC & LC CIRCUITS a more general form that was initially derived Alternating Direct from the Biot-Savart Law. In situations with Current Current high symmetry, Ampère's Law simplifies the Amount of Safe to transfer calculation of the magnetic field. energy over longer city  Consistent Results: Both Biot-Savart and that can distances and Ampère's Law give consistent results when be carried can provide applied correctly to the same physical more power. Frequency The frequency The frequency situation. Ampère's Law is essentially a more of alternating of direct current generalized and often more convenient form current is 50Hz is zero. for certain cases. or 60Hz depending upon the ELECTROSTATIC F. VS INDUCED E. FIELDS country. Electrostatic Fields Induced E. Fields Direction It reverses its It flows in one  Start at positive  Non direction while direction in the charges and ends electrostatic flowing in a circuit. at negative fields circuit. Current It is the current It is the current charges.  Do not start/end of magnitude of constant  Conservative at a charge; varying with magnitude. field they just go time around loops. Flow of Electrons keep Electrons move  Non Electrons switching steadily in one conservative directions - direction or field. forward and 'forward'.  It only requires backward. varying Obtained A.C Generator Cell or Battery magnetic field. from and mains.  With or without free electrons. INDUCTANCE  Inductor – The device placed in a circuit to oppose a change in current; that is, to maintain, and regulate, a steady current in that section of the circuit (solenoid).  If a current increases through a solenoid, magnetic field increases.  Changing magnetic field = emf INDUCED ELECTRIC FIELD  Lenz’s law – emf opposes change that  The changing magnetic field creates an induced it electric field on the loop, in the same direction  Slows down the increase of current (back as the current. The electric field directed emf) tangent toward the loop exerts an electric  Back emf – prevents the force on the charges all along the loop. increase/decrease in current or  This induced electric field is different from the fluctuations. electrostatic field.  Current (steady) constant magnetic field = no emf 𝑵𝝓𝑩 𝑳= 𝒊 fau GENERAL PHYSICS: MAGNETISM INTRODUCTION | MAGNETIC FIELD, FORCE, FLUX | FARADAY’S LAW | LENZ’S LAW | BIOT-SAVART LAW | AMPERE’S LAW | INDUCED ELECTRIC FIELD | AC & LC CIRCUITS SHS | STEM | SECOND SEMESTER | 2024 INDUCTANCE, L  Inductance is the tendency of an electrical conductor to oppose a change in the electric current flowing through it.  ↑ inductance, ↑ ability to oppose change  Unit: Henry, H (Weber/Ampere)  Self – inductance  Mutual Inductance CIRCUITS SELF – INDUCTANCE  Inductor will be able to oppose a change in the electric current flowing by itself. 𝝁𝒐 𝑵𝟐 𝑨 𝑳= 𝒍  𝐿 – inductance, Henry (Weber/Ampere)  𝜇𝑜 – diameter of the coil in meters, permeability of free space (4𝜋𝑥10−7 𝑇𝑚/ 𝐴)  A capacitor charged to a potential difference  𝑁 – number of turns in the coil Vc is connected to an inductor. Charges will  𝐴 – cross-sectional area migrate from the positive plate to the negative  𝑙 – length of the coil in meters plate. A current (I) starts to flow across the inductor. As current increases, the inductor MUTUAL INDUCTANCE generates an opposing current (IIND),  Need another inductor to induce back emf. slowing it down. 𝑵𝟐 𝝓𝑩𝟐 𝑵𝟏 𝝓𝑩𝟏 𝑴= = 𝒊𝟏 𝒊𝟐  SI Unit: H (Weber/Ampere)  For the induced emfs: 𝒅𝒊𝟏 𝒅𝒊𝟐 𝜺𝟐 = −𝑴 and 𝜺𝟏 = −𝑴 𝒅𝒕 𝒅𝒕  SI Unit: V Transformers  Device used to step up or step down voltage.  The charges from the capacitor will be depleted at some point and current (I) = 0 A.  Step-down transformer – the output The induced emf (VL) in the inductor will voltage is lower than the input voltage. reverse direction. The inductor then will  Step-up transformer – the output generate an induced current (IIND) to slow voltage is higher than the input voltage. down the decrease of the current, reversing  Lifeblood: Mutual Inductance the polarity of the capacitor. LC CIRCUITS  Then, the capacitor will be charged again, but  Sometimes called oscillatory circuit. with different polarity. It will drain again and  Contains an inductor and a capacitor. the process will start again.  When a charged capacitor is connected to an inductor, energy oscillates from electrical to magnetic, and back to electrical, and so on. fau GENERAL PHYSICS: MAGNETISM INTRODUCTION | MAGNETIC FIELD, FORCE, FLUX | FARADAY’S LAW | LENZ’S LAW | BIOT-SAVART LAW | AMPERE’S LAW | INDUCED ELECTRIC FIELD | AC & LC CIRCUITS SHS | STEM | SECOND SEMESTER | 2024 Applications AC Circuit LC Circuit  Transformers  When the knob of the radio is moved (i.e., we are turning the radio)  Capacitance of the capacitor is changing by changing the knob.  Capacitance keeps on changing until the resonant frequency becomes equal to the frequency of the channel which we want to hear. fau