1 / 20

Lesson 63 – Motion of Charged Particles in Magnetic Fields Part 2

Lesson 63 – Motion of Charged Particles in Magnetic Fields Part 2. By: Fernando Morales, Brian Cox’s Brother October 25, 2013. Magnetic Force on a Moving Charge. The unit of magnetic field strength is the Tesla, T. F B = qvBsin . [N] = [C][m/s][T].  1 T = 1 Nm/Cs. EXAMPLE.

zlhna
Download Presentation

Lesson 63 – Motion of Charged Particles in Magnetic Fields Part 2

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Lesson 63 – Motion of Charged Particles in Magnetic Fields Part 2 By: Fernando Morales, Brian Cox’s Brother October 25, 2013

  2. Magnetic Force on a Moving Charge The unit of magnetic field strength is the Tesla, T FB = qvBsin [N] = [C][m/s][T]  1 T = 1 Nm/Cs EXAMPLE Calculations (a) find F, if v = 105 m/s, q = 10 nC, B = 500 mT,  = 90° (b) Find v, if F = 10 nN, B = 200 mT,  = 60°, q = 100 mC (c) find  if F = 200 mN, B = 2.5 T, q = 200 nC, v = 2 (106) m/s (d) Find F, if v = 5(106) m/s, B = 2 T, q = 2 nC,  = 0 (e) under what condition does a charge experience no force when placed in a B field? (a) F = qvBsin = 10(10-9)(105)(500(10-3))sin90 = 500 N (b) v = F/qBsin = 10(10-9)/(100(10-3)200(10-3) sin60) = 0.577 (10-6) m/s Solutions (c)  = arcsin(F/qvB) = 11.5 º (e) v || B or v = 0 (d) 0

  3. F v B Directions of magnetic force, velocity & magnetic field FB, & F v Directions are given by Right Hand Rule Thumb  velocity Fingers  B Palm  Force, for +ve

  4. B, F, v directions In each diagram use the RHR to find the missing direction of the F, field or velocity or to find the sign of the charge EXAMPLE F v B X20 – 0a x x x x x x x x x x x x x x x 1, F 3, v ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● 5, B ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● x x x x x x x x x x x x x x x 2, F 4, q ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● 6, v ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● FB = qvBsin -ve +ve Direction is given by Right Hand Rule B Thumb  velocity Fingers  B F Palm  Force, for +ve v

  5. B, F, v directions In each diagram use the RHR to find the missing direction of the F, field or velocity or to find the sign of the charge EXAMPLE F v B X20 – 0b 1, F 3, F ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● 5, F x ● No force, v // B No force, v anti // B 2, F ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● 4, v 6, F ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● x FB = qvBsin -ve +ve Direction is given by Right Hand Rule B Thumb  velocity Fingers  B F Palm  Force, for +ve v

  6. EXAMPLE B, F, v directions In each diagram use the RHR to find the missing direction of the F, field or velocity or to find the sign of the charge F v B X20 – 0c ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● 1, B 3, F ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● 5, B ● 2, q 4, v ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● 6, q ● ● FB = qvBsin -ve +ve Direction is given by Right Hand Rule B Thumb  velocity Fingers  B F Palm  Force, for +ve v

  7. Path of a Moving Charge in a Magnetic Field F mv² v r mv qB FB = qvBsin B Direction is given by Right Hand Rule Thumb  velocity Fingers  B Palm  Force, for +ve The force on a moving charge in a magnetic field is  to v and B  a circular path ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● Both +ve and –ve charges move in circles but in the opposite sense = qvB r =

  8. EXAMPLE mv² r mv qB F v B Electron’s Path in a uniform B field An electron has a speed of 2.0 (107) m∕s in a plane perpendicular to a 0.010 T B field. Describe the path. Qe = -1.6 (10-19) C X20 – 4 Solution The electron moves at constant speed in a curved path (the force is  to the velocity so there is NO change in speed only direction). The radius of curvature is found from F = ma and a = v²∕r .  = 90 Since F = qvB = ma qvB = r = r = 1.1 (10-2) m FB = qvBsin Direction is given by Right Hand Rule Thumb  velocity Fingers  B Palm  Force, for +ve

  9. EXAMPLE F v B What is the path of a charged particle in a uniform magnetic field if the velocity is not  to the field? X20 – 5 Solution The velocity vector has components // to B and  to B. The // component results in no force and is unchanged by B. The  component produces a circular path. Combining the two results produces a helix FB = qvBsin Direction is given by Right Hand Rule Thumb  velocity Path of charged particle in B field Fingers  B Palm  Force, for +ve

  10. A proton gun shoots out a single proton with a speed of 3640000 m/s targeting a stationary alpha particle with a charge of 2e. The distance between the target and the gun is 2.00 m. How close will the proton come to the alpha particle before coming to an instantaneous stop? Hint: Look up the mass of proton

  11. EXAMPLE Aurorae borealis – the Northern Lights Charged particles approach the Earth from the Sun (the “solar wind”) and are drawn to the poles creating bright lights in the sky. Why at the poles? X20 – 6 Solution As the particles approach earth they have velocity components // to and  to Earth’s B field. This causes the particles spiral around the field lines heading to the Poles, where the lines enter the Earth. The high concentration of charged particles ionizes the air in the North, and this produces light as the electrons recombine with the ions.

  12. X2 - 3 EXAMPLE What fields surround a moving electric charge? An electric field due to the charge of the electron, a magnetic field as the charge is in motion and a gravitational field as the electron has mass

  13. X2 - 3 EXAMPLE If 2 iron bars are placed close together and attract each other regardless of which ends are placed together, are both magnets? One of the bars is a magnet and attracts the other; if both were magnets then when the 2 N and 2S ends were proximal (close) they would repel each other

  14. X2 - 3 EXAMPLE Can you start an electron at rest in motion with a magnetic field? With an electric field? You cannot move an electron at rest with a magnetic field (magnetic force depends on the velocity of the charge, 0 velocity  0 force and acceleration). Of course, an electric field can move an electron.

  15. X2 - 3 EXAMPLE A charged particle is moving in a circle in a magnetic field, B. An electric field E pointed in the same direction as B is turned on. Describe the path the particle follows. The electric field drives the particle on the axis of the field (|| or anti-||), while the particle continues to do circles due to its velocity  B. The net effect will be a stretched (due to acceleration) helix, or screw-like path along the axis of the electric field

  16. Required Before Next Class • Chapter 8 Self Quiz # 8, 10, 16 • Chapter 8 Review # 1, 7, 13, 16, 17, 20, 21, 22, 23, 24, 25, 26, 27, 28, 37, 39, 40, 42, 50, 65, 66

More Related