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Physics Project 6th six weeks

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• If we place a current-carrying wire in an external magnetic field, it will experience a force.
• ### M and E

1. 1. JONATHAN ARNOLD ALEXA P. DELGADO Magnetic Fields and Electromagnetism
2. 2. Magnetic Fields <ul><li>electromagnet: a magnet with a field produced by an electric current </li></ul><ul><li>law of poles: like poles repel each other and unlike poles attract </li></ul><ul><li>magnetic domain: cluster of magnetically aligned atoms </li></ul><ul><li>magnetic field: the space around a magnet in which another magnet or moving charge will experience a force </li></ul><ul><li>mass spectrometer: a device which uses forces acting on charged particles moving a magnetic field and the resulting path of the particles to determine the relative masses of the charged particles </li></ul><ul><li>right-hand rules: used to find the magnetic field around a current-carrying wire or the force acting on a wire or charge in a magnetic field </li></ul><ul><li>Solenoid: a long coil of wire in the shape of a helix; when current is passed through a solenoid it produces a magnetic field similar to a bar magnet </li></ul>
3. 3. Catch up on Symbols <ul><li>B = magnetic field </li></ul><ul><li>F B = magnetic force </li></ul><ul><li>q = charge </li></ul><ul><li>v = speed or velocity of a charge </li></ul><ul><li>θ = angle between the velocity </li></ul><ul><li>r = radius of path of a charge </li></ul><ul><li>m = mass </li></ul><ul><li>I = current </li></ul><ul><li>L = length of wire in a magnetic field </li></ul><ul><li>μ 0 = permeability constant </li></ul>
4. 4. <ul><li>Right-hand for force on a moving charge: </li></ul><ul><li>The equation for finding the force on a charge moving through a magnetic field is… </li></ul><ul><ul><li>Step one: Place your fingers in the direction of the magnetic field (north to south) </li></ul></ul><ul><ul><li>Step two: Your thumb should be in the direction of the velocity of a moving charge (or current in a wire), </li></ul></ul><ul><ul><li>Step three: Then you will find the magnetic force on the charge (or wire), because it will come out of the direction of your palm.  </li></ul></ul><ul><ul><li>F = qvBsin  </li></ul></ul><ul><ul><ul><li>where q is the charge in Coulombs </li></ul></ul></ul><ul><ul><ul><li>v is the velocity in m/s </li></ul></ul></ul><ul><ul><ul><li>B is the magnetic field in Teslas </li></ul></ul></ul><ul><ul><ul><li> is the angle between the velocity and the magnetic field </li></ul></ul></ul><ul><ul><ul><ul><li>If the angle is 90  , the equation becomes F = qvB. </li></ul></ul></ul></ul>Forces on moving charges in magnetic fields
5. 5. Forces on current-carrying wires in magnetic Fields <ul><li>Since a current-carrying wire creates a magnetic field around itself every current-carrying wire is a magnet. </li></ul><ul><li>Once again, the direction of the force acting on the wire is given by the right-hand rule that we previously explained. </li></ul><ul><ul><li>Again, you would use your left hand to find the direction of the magnetic force if you were given electron flow instead of conventional current. </li></ul></ul><ul><li>The equation for finding the force on a current-carrying wire in a magnetic field is </li></ul><ul><li>F = ILBsin  </li></ul><ul><li>where I is the current in the wire, L is the length of wire which is in the magnetic field, B is the magnetic field, and  is the angle between the length of wire and the magnetic field. </li></ul><ul><ul><li>If the angle is 90  , the equation becomes simply F = ILB. </li></ul></ul>
6. 6. <ul><li>A current-carrying wire creates… </li></ul><ul><li>Right-hand Rule # 2 for the magnetic field around a current-carrying wire: </li></ul><ul><li>a magnetic field around itself. </li></ul><ul><li>Magnetic fields are produced by moving charges. </li></ul><ul><ul><li>This is why all atoms are tiny magnets, since the electrons around the nucleus of the atom are moving charges and are therefore magnetic. </li></ul></ul><ul><li>The magnetic field due to a current-carrying wire circulates around the wire in a direction can be found by another right-hand rule. </li></ul><ul><li>Curl your fingers into a half-circle around the wire, they point in the direction of the magnetic field, B Point your thumb in the direction of the conventional current I. </li></ul><ul><li>In determining the direction of a magnetic field due to the flow of electrons in a wire, we would use the left hand instead of the right hand.   </li></ul>Fields of long current-carrying wires
7. 7. Ampere’s law <ul><li>Ampere’s law can be used to find the magnetic field due to a steady current flowing in a very long straight wire. </li></ul>
8. 8. Biot-Savart law <ul><li>The magnetic equivalent of Coulomb's law is the Biot-Savart law for the magnetic field produced by a short segment of wire , carrying current I : </li></ul><ul><li>where the direction of is in the direction of the current and where the vector points from the short segment of current to the observation point where we are to compute the magnetic field. Since current must flow in a circuit, integration is always required to find the total magnetic field at any point. The constant is chosen so that when the current is in amps and the distances are in meters, the magnetic field is correctly given in units of tesla. Its value in our SI units is exactly </li></ul><ul><li>Infinitely lonf wire: </li></ul><ul><li>Circular loop: </li></ul><ul><li>Long thick wire: </li></ul><ul><li>Long Solenoid: </li></ul><ul><li>Toroid: </li></ul>
9. 9. Example UNO <ul><li>Q: A wire carrying 4.0 A of current is placed in a uniform magnetic field of strength 2.0 T as shown in the diagram that should be located on the white board by now… a) use the right-hand rule to determine the direction of the force on the wire b) Determine the magnitude of the force on a 0.010 meter section of the wire. </li></ul><ul><li>A: a)The current is directed toward the top of the page… wondering why? I shall explain….. </li></ul><ul><li>b) F= I ℓ B sin Ø </li></ul><ul><li> = (4.0 A) (0.010 m) (2.0 T)(sin 90 ̊) </li></ul><ul><li> F= 0.080 N </li></ul>
10. 10. Example #2 <ul><li>Q: A alpha particle has a mass of 6.68 10 ⁻²⁷ kg and charge 3.2 10⁻¹⁹ c. The particle is traveling at 3.0 10⁶ m/s and enters a magnetic 0.10 T. The magnetic field is directed at right angles to the direction of motion of the particle. Determine the radius of the circle in which the particle travels. </li></ul><ul><li>Look at the board for picture… </li></ul><ul><li>A: net F= mv ²/r but </li></ul><ul><li>a c = v ² /r </li></ul><ul><li>F magnetic = mv ² /r </li></ul><ul><li>qvBsinØ= mv ² /r </li></ul><ul><li>But! Ø=90 ̊ and sin90 ̊=1 </li></ul><ul><li>and solving for r gives.. </li></ul><ul><li>r= (mv)/(qB) </li></ul><ul><li>= [( 6.68 10 ⁻²⁷ kg)(3.0 10⁶ m/s)]/[(3.2 10⁻¹⁹ c) (0.10 T)] </li></ul><ul><li>r= 0.63m </li></ul>
11. 11. Example… THREE!!! <ul><li>Q: What kind of field or fields surround a moving electric charge? </li></ul><ul><li>A: A moving electric charge produces a magnetic field. A electric charge, whether moving or not, has an electric field in the region surrounding the charge. However, the electric charge has mass and must be surrounded by a gravitational field. Therefore, and electric field, a magnetic field, and a gravitational field all surround a moving electric charge. </li></ul>
12. 12. Electromagnetism Symbols <ul><li>ε = emf (voltage) induced by electromagnetic induction </li></ul><ul><li>v = relative speed between a conductor and a magnetic field </li></ul><ul><li>B = magnetic field </li></ul><ul><li>L = length of a conductor in a magnetic field </li></ul><ul><li>I = current </li></ul><ul><li>R = resistance </li></ul><ul><li>Φ = magnetic flux </li></ul><ul><li>A = area through which the flux is passing </li></ul><ul><li>N = number of loops in a coil of wire </li></ul><ul><li>V S = voltage in the secondary coil of a transformer </li></ul><ul><li>V P = voltage in the primary coil of a transformer </li></ul><ul><li>N S = number of loops in the secondary coil of wire in a transformer </li></ul><ul><li>N P = number of loops in the primary coil of wire in a transformer </li></ul>
13. 13. Electromagnetic induction <ul><li>electromagnetic induction: an electric current being produced by moving a magnet through a coil of wire, or, equivalently, by moving a wire through a magnetic field </li></ul><ul><li>Faraday’s law: the magnitude of the induced emf in a coil depends on the number of turns in the loop and the rate of chage of flux through the loop </li></ul><ul><li>Lenz’s law: There is an induced current in a closed conducting loop if and only if the magnetic flux through the loop is changing. The direction of the induced current is such that the induced magnetic field always opposes the change in the flux. </li></ul><ul><ul><li>The changing magnetic field in the coil due to the change in current induces a current in the metal ring. Lenz’s Law states that the magnetic field in the ring opposes the magnetic field of the coil and forces the ring to fly away. </li></ul></ul>
14. 14. Inductance <ul><li>Current rise in a LR circuit: The current (I) in an LR circuit is related to the emf of the source, the resistance the time constant, and time. </li></ul><ul><li>The inductive time constant of an L circuit. </li></ul><ul><li>Current decay in an LC circuit: The current (I) in an LR decay circuit depends on the emf of the source the resistance (R), the time constant (T), and time (t) </li></ul>
15. 15. Maxwell’s Equations <ul><li>Consist of four basic equations that describe all electric and magnetic phenomena. </li></ul><ul><ul><li>Gauss’ law for electricity is a generalized form of Coulomb’s law and relates electric charge to the electric field that the charge produces </li></ul></ul><ul><ul><li>Gauss’ law of magnetism describes magnetic fields and predicts that magnetic fields are continuous, they have no beginning or ending point. The equation predicts that there are no magnetic monopoles </li></ul></ul><ul><ul><li>Faraday’s law of induction describes the production of an electric field by a changing magnetic field. </li></ul></ul><ul><ul><li>Maxwell’s extension of Ampere’s law describes the magnetic field produced by a changing electric field or by an electric current </li></ul></ul>
16. 16. Examples! problem1 <ul><li>A radar pulse travels from the Earth to the Moon and back in 2.60 seconds. Calculate the distance from the Earth to the Moon in a) meters and b) miles. </li></ul><ul><li>Solution (a): A radar pulse is an electromagnetic wave which travels at 3.00x10^8 m/s. The time required for the pulse to travel from the Earth to the Moon is equal to (1/2)(2.60s) = 1.30s. Since it travels in a straight line at a constant speed, the distance traveled is determined by: </li></ul><ul><li>d = ct = (3.00x10^8 m/s)(1.3s) </li></ul><ul><li>So then d = 3.90x10^8 m </li></ul><ul><li>Solution (b): 1.000 miles = 1,609 meters, so </li></ul><ul><li>d = (3.90x10^8 m) x (1.000 mile/ 1,609 meters) which equals: </li></ul><ul><li>d = 2.42x10^5 miles </li></ul>
17. 17. Examples! Problem2 <ul><li>Determine the frequency of yellow light of wavelength 6.00x10^-7 meters. </li></ul><ul><li>Solution: The frequency, wavelength, and velocity of light are related by the equation c = (frequency) x (wavelength), where c refers to the speed of light in a vacuum. </li></ul><ul><li>f = (c)/(wavelength) = </li></ul><ul><li>(3.00x10^8m/s)/(6.00x10^-7 m) so, </li></ul><ul><li>f = 5.00x10^14 Hz </li></ul>
18. 18. Examples! problem3 <ul><li>The energy density of an EM wave is (5.00x10^-5) J/m^3. Determine the peak magnitude of the electric field strength. </li></ul><ul><li>Solution: The energy density is related to the electric field strength by the equation: u = € ο E^2, </li></ul><ul><li>But E constant^2 = E ο ^2/2 therefore, </li></ul><ul><li>u = ½ € ο Eo^2 rearranging gives: </li></ul><ul><li>Eo = (2u/€ ο )^½ which equals: </li></ul><ul><li>Eo = [(2)(5.00x10^-5 J/m^3)/(8.85x10^-12 C^2/N m^2)] ^½ which makes: </li></ul><ul><li>Eo = 3.36x10^3 N/C </li></ul>
19. 19. Hope this made sense and you all enjoyed!!!