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MSTC Physics C

MSTC Physics C. Study Guide Chapter 22 Sections 5-6. Current-Carrying Wire. Suppose a straight wire is placed between the poles of a magnet When I flows in wire a force is exerted on it which is perpendicular to both B field and direction of I So wire is forced either up or down. N. S.

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MSTC Physics C

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  1. MSTC Physics C Study Guide Chapter 22 Sections 5-6

  2. Current-Carrying Wire • Suppose a straight wire is placed between the poles of a magnet • When I flows in wire a force is exerted on it which is perpendicular to both B field and direction of I • So wire is forced either up or down N S

  3. Current-Carrying Wire • Direction of force on wire is determined by right hand rule - point fingers in direction of current and bend fingers in direction of B field; thumb points in direction of force N S

  4. Current-Carrying Wire • Force on a particle is given by F = qv x B • Consider a wire of length l and cross sectional area A carrying a current I • For all particles F = (qvx B)nAl where Al is volume and n is the number of charges per volume N S

  5. Current-Carrying Wire • Recall I = nqvA • So F = Il x B = IlBsinΘ • If B field isn’t uniform or wire isn’t straight dF = Idl x B for each piece • So F = ∫ Idl x B N S

  6. Sample Problem • A rectangular loop of wire hangs vertically. A magnetic field is directed horizontally, perpendicular to the wire, and points out of the page at all points. The magnetic field is very nearly uniform along the horizontal portion of the wire which is near the center of the large magnet producing the field; the magnet is aligned symmetrically so that the non-uniformity of B along the section of vertical wire is the same for both vertical lengths; the top portion of the wire loop is entirely free of the field. The loop hangs from a balance which measures a downward force of 0.0348 N when the wire carries a current of 0.245 A. What is the magnitude of the magnetic field at the center of the magnet?

  7. Sample Problem • A rigid wire, carrying a current I, consists of a semicircle of radius R and two straight portions. The wire lies in a plane perpendicular to a uniform magnetic field. The straight portions each have length l within the field. Determine the net force on the wire due to the magnetic field.

  8. Closed Loop x xxxxxxxx x xxxxxxxx x xxxxxxxx • What is the force on a closed loop in a uniform B field? F = ∫ Idl x B but ∫ dl = 0 since start and stop at the same point so F = 0 ! For any closed loop in a uniform B field, F = 0

  9. Torque on a Loop • Consider a loop of length a and width b • Loop carries current I • Horizontal B field makes an angle Θwith normal to plane surface of loop • Loop free to rotate about vertical axis

  10. Torque on a Loop • Magnetic force on top and bottom section of loop cancel (equal magnitude and length, opposite direction) • For each vertical section F = Il x B = IlBsinΘ = IaB • for left section force is into page • For right section force is out of page Don’t cancel since don’t act along same line. I

  11. Torque on a Loop • Torque is produced Τ = F x l = FlsinΘ ΣT = IaB(b/2)sinΘ = IaBbsinΘ let ab = area of loop, A ΣT = IABsinΘ I

  12. Torque on a Loop • If loop contains N coils ΣT = NIABsinΘ Note: T is max when B is parallel to plane of loop Note: T is min when B is perpendicular to plane of loop I

  13. Magnetic Moment • Vector quantity whose direction is perpendicular to the plane of the loop • μ = NIA • So T = NIA x B = μ x B

  14. Positive particle perpendicular to B field • Does a magnetic force do work on a charge? No • What does a magnetic force do to a moving charge? can change direction but not its speed of kinetic energy • A positive particle moving in a B field with its initial speed perpendicular to field will move in a circle whose plane is perpendicular to B

  15. Positive particle perpendicular to B field x x x x x x x vo x x x x x x x x x x x x x x • Here rotates counterclockwise F = mv2/r qvBsinΘ = mv2/r r = mv/qB

  16. Positive particle perpendicular to B field x xxxxxx vo x xxxxxx x xxxxxx • Since v = 2πr / T • Period would be T = 2πr/v = 2πm/qB • Frequency would be f = 1/T = qB/2πm • Angular velocity is ω = v/r = qB/m Note: T, f, ω don’t depend on particle’s velocity or radius

  17. Positive particle with an angle to B field y x z • Here its path will be a helix • No Fx = 0 and no ax so vx is constant • Since F = qv x B = qvBsinΘ so the F will cause the v to change in the y and z direction vo

  18. Sample Problem • A current of 17 mA is maintained in a single circular loop of 2 m circumference. An external magnetic field of 0.8 T is directed parallel to the plane of the loop. A) Calculate the magnetic moment of the current loop. B) What is the magnitude of the torque exerted on the loop by the magnetic field?

  19. Sample Problem • A circular coil of 100 turns has a radius of 0.025 m and carries a current of 0.1 A while in a uniform external magnetic field of 1.5 T. How much work must be done to rotate the coil from a position where the magnetic moment is parallel to the field to a position where the magnetic moment is opposite the field?

  20. Sample Problem • A singly charged positive ion has a mass of 3.2 x 10-26 kg. After being accelerated through a potential difference of 833 V, the ion enters a magnetic field of 0.92 T along a direction perpendicular to the direction of the field. Calculate the radius of the path of the ion in the field.

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