1 / 26

Magnetism

Magnetism. Forces caused by Magnetic Fields. Magnetic Fields. Measured in Teslas (T) or Gauss (G) 10,000 G = 1 T Moving charge causes magnetic fields Remember the domain theory If moving charge causes magnetic fields, can a magnetic field affect a moving charge?. Force on a Moving Charge.

eldora
Download Presentation

Magnetism

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. Magnetism Forces caused by Magnetic Fields

  2. Magnetic Fields • Measured in Teslas (T) or Gauss (G) • 10,000 G = 1 T • Moving charge causes magnetic fields • Remember the domain theory • If moving charge causes magnetic fields, can a magnetic field affect a moving charge?

  3. Force on a Moving Charge • A moving charge experiences a force in a magnetic field • F = qv x B • The magnitude of the force is qvB sinθ • The direction is determined by the Right Hand Rule

  4. Apply RHR B F X B v Determine the direction of the velocity of the charge. Determine the direction of the force on the charge if the charge is a proton.

  5. Apply RHR v v B F X B F

  6. Force on a Current Carrying Wire • A current carrying wire experiences a force in a magnetic field • dF = I dL x dB • F = IL x B • The magnitude of the force is IlB sinθ • The direction is determined by the RHR

  7. Problem 1 • A wire carrying a current of 2.0 A lies along the x-axis. The current flows in the positive x direction. A magnetic field of 1.2 T is parallel to the xy plane and makes an angle of 30º with the x-axis (pointing into the 1st quadrant.) What is the force on a segment of wire 0.40 m long?

  8. Solution • F = IlB sinθ • F = (2.0 A)(0.40 m)(1.2 T) sin 30º • F = 0.48 N k

  9. Problem 2 • A wire carrying current of 1.5 A lies in a horizontal surface in the xy plane. One end of the wire is at the origin and the other is at (3m,4m). The wire follows an erratic path from one end to the other. A magnetic field of 0.15T directed vertically downward is present. What is the magnetic force acting on the wire?

  10. Solution • F = ILB sin θ • F = 0.15T (1.5A) (3i-4j) • F = 0.225 √(32 – 42) • F = 1.13 N • This is the magnitude only!

  11. Homework • Problems Ch. 29 • #1-5 and #33-38

  12. F = qv x B • Magnetic force on a charge at rest is zero. • The magnetic force on a charge moving parallel or anti-parallel to the magnetic field is zero. • Magnetic force, when non zero is perpendicular to velocity and magnetic field. Thus it contributes only to a radial acceleration and never performs work on an object or changes its speed (it can only change its direction).

  13. Continued • Magnetic force is proportional to the charge of a point charge and thus exerts forces only on charged objects. • The direction can be obtained by the right-hand rule (positive charge only—reverse the direction of the velocity for a negative charge).

  14. T = N/(Cm/s) T = N/Am Tesla

  15. Change of direction • F = qvB • This force causes circular motion • F = mv2/r • mv2/r = qvB • B = mv/qr • Or m/q = Br/v (mass to charge ratio) • Mass spectrometry allows us to find the mass to charge ratio

  16. Three types of Motion • When a charged particle moves into a magnetic field three types of motion can occur: • Move parallel to B and experience no force • Move perpendicular to B supplying a centripetal force • Move at some angle to B and have components in the parallel and perpendicular direction producing helical or corkscrew pattern of motion

  17. F = qv x B F = e(Nv0) x B N = # charges N = n(volume) = nlA Where l is the length of the wire and A is the cross-sectional area F = nlA evo x B I = JA I = nevDA so F = Il x B dF = i dl x B F = ∫idl x B= i∫dl x B Force on a Current Carrying Wire

  18. Force on a Semicircle of Wire • A wire bent into a semicircle of radius R forms a closed circuit and carries a current I. The wire lies in the xy plane and a uniform magnetic field is directed along the positive y axis. Find the magnitude and direction of the magnetic force acting on the straight portion of the wire and on the curved portion.

  19. F1 acting on the straight portion has a magnitude of F1 = IlBsinθ = I(2R)B sin 90° F1 = 2IRB Direction out of the page or +z axis Answer B

  20. F2 acting on the curved part—We must first write an equation for dF2 on the length ds dF2 = I ds X B = IB sin θ ds dF2 = IB sin θ ds s = R θ so ds = r d θ Continued B

  21. θ = 0 to θ = π dF2 = IRB sin θ d θ F2 = IRB ∫ sin θ d θ From 0 to π F2 = IRB (-cos θ) from 0 to π = -IRB (cos π – cos 0) = -IRB (-1-1) = 2IRB Direction: into the page or –z axis Continued B

  22. Torque = F x r Torque max = Frsinθ Torque max = F2 (b/2)sin 90 + F4(b/2)sin 90 = IaB(b/2) +IaB(b/2) = (IAB)/2 + (IAB)/2 = IAB Where A = ab Torque on a Current Carrying Loop I B 1 a 4 2 b 3

  23. Ends or part 1 and 3 Torque = F1 (a/2)sinθ + F3(a/2)sinθ Torque = IbB(a/2)sinθ + IbB(a/2)sinθ = IabBsinθ = IAB sinθ Continued I B 1 a 4 2 b 3

  24. Torque = I(A x B) μ = IA Torque = μ x B And U = - μ dot B A = Area Vector Magnetic moment

  25. Problem • A rectangular coil of dimensions 5.40 cm x 8.50 cm consists of 25 turns of wire and carries a current of 15.0 mA. A 0.350T magnetic field is applied parallel to the plane of the loop. A) Calculate the magnitude of its magnetic dipole moment. B) What is the magnitude of the torque acting on the loop?

  26. Answer • μ = NIA = 25 (15 x 10-3 A)(0.0540 m)(0.0850 m) = 1.72 x 10-3 Am2 • Torque = 1.72 x 10-3 Am2 (0.350T) = 6.02x10-4 Nm

More Related