1 Chapter 17 Magnetic Field and Magnetic Forces. 2 Magnetism S N South North South magnetic pole South geographic pole Earth’s magnetic field North magnetic

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Text of 1 Chapter 17 Magnetic Field and Magnetic Forces. 2 Magnetism S N South North South magnetic pole...

  • *Chapter 17Magnetic Fieldand Magnetic Forces

  • *MagnetismOpposite poles : attract each other

    Like poles: repel each other The needle of a compass aligns with the magnetic field Earth is a magnetic. The axis of earths magnetic is not parallel to its geographic axisMagnetic declination(Rotation axis)(a vector field)

  • *

  • *Magnetic FieldIn addition to the electric field, a moving charge or a current in space can create a magnetic field.An electric force (F = Q0E) will exert on other charge (Q) present in the electric field (E). Similarly, the magnetic field also exerts a magnetic force on other moving charge or current present in the magnetic field.Oersteds Experiment

  • *Magnitude of magnetic force is proportional to:

    magnitude of the chargemagnitude or strength of the magnetic fieldvelocity of the moving particle (for electric force, it is the same no matter the charge is moving or not) or the component of velocity perpendicular to the field.

    A charged particle at rest will have no magnetic force.

    The direction of magnetic force (F) is not the same as the direction of magnetic field (B). Instead, the magnetic force is always perpendicular to both direction of magnetic field (B) and the velocity (v).Direction of B: the north pole direction of a compass needle.

    For a magnet, the direction of B is pointing out of its north pole and into its south pole.

  • *Magnetic force on a moving charged particle:where:

    F : magnetic force [N]Q : magnitude of charge [C]v : velocity of the charge [m/s]B : magnetic field [T or Ns/Cm or N/Am (A: ampere)]

    1 N/Am = 1 tesla = 1 T [Nikola Tesla (1857 1943)]

  • *For a negative charge, the force is opposite to the case of the positive charge.

  • *Magnetic FluxTotal magnetic flux through a surface A:Sum of magnetic flux thru areas of all elementsWhere:

    A: magnetic flux (a scalar) [weber (Wb)]B : magnetic field [T]A : surface area [m2]1 Wb = 1 T m2 = 1 Nm / ATotal magnetic flux through a closed surface = 0Gausss Law for MagnetismMagnetic field B is also called magnetic flux density In = Out

  • *Motion of Charged Particles in a Magnetic FieldA charge particle under the action of a magnetic field only moves with a constant speed. The motion is determined by Newtons laws of motion.Circular motion of a positive charge in a uniform magnetic field (B):x denotes that the magnetic field is pointing into the planem : mass of the particlev : constant velocity R : radius of the circular orbit

  • *Magnetic Force on a ConductorMagnetic force on a straight wire:Where:

    F: magnetic forceI : total currentL : length of the wire segmentMagnetic force on an infinitesimal wire (not straight wire):Divide the wire into infinitesimal straight line

  • SOURCES OF MAGNETIC FIELD *Source point is referred to the location of a charge (Q) moving with a constant velocity (v) in a magnetic field.Field point is referred to the location or point where the field is to be determined, e.g. location of point k. Magnetic field of a point charge moving with a constant velocity:where:B : magnetic field Q: point chargev: velocity of the charger: distance from the charge to the field point0 = 4 10-7 Ns2/C21 Ns2/C2 = 1 Wb/Am = 1 Tm/A = 1 n/A2

  • *Applying the principle of superposition, the magnetic fields of a number of moving charges can be calculated. Total magnetic field due to a number of moving charges = vector sum of the electric fields due to the individual chargesLaw of Biot and Savart for Magnetic Field of a Current Element (B):wheredL : represents a short segment of a current-carrying conductorI: current in the segmentnQ : total chargesvd : drifting velocityA : cross-section area of segment

  • *Magnetic field of a straight current-carrying conductor:

  • *Example 17.1:The figure shows an end view of two parallel wires carrying the same current I in opposite directions. Determine the magnitude and direction of magnetic flux B at point A.Solution:Use principle of superposition of magnetic fields:

    Btotal = B1 + B2

    Point A is closer to wire 1 than to wire 2, the field magnitude B1 > B2Use right hand rule, B1 is in the y-direction and B2 is in the + y-direction. As B1 > B2, Btotal is in the y-direction and the magnitude is:x : into the plane : out of the plane

  • *Ampere LawThe line integral of magnetic field intensity around a single closed path is equal to the algebraic sum of currents enclosed.Integration path not enclosing the conductorIntegration path enclosing the conductor

  • *Example 17.2A long, straight wire with radius of a, and the wire carries a current I0, which is distributed uniformly over its cross section. Find the magnetic field both inside and outside the wire.

  • *Solution:By applying Amperes Law,BdL = BdL = BdL = B(2r)For r a, (inside the wire)I = I0=> I = I0where I0 is the total current over the cross-section of the wire. B(2r) = 0( ) I0So, we have B = ( r a )For r > a, (outside the wire)( r > a )B =

    r2a2

    r2 a2

    r2 a2

    0 I 0 2

    r a2

    0 I 0 2r

  • *In-Class Exercise 17.1: (Magnetic field of a circular current loop)Determine the magnitude and direction of the magnetic field at point P due to the current in the semicircular section of wire shown in figureSolution:

    There is no magnetic field at the center of the loop from the straight sections. The magnetic field from the semicircle is just half that of a complete loop:Into the page.

  • *Magnetic Field of a Circular Current Loop

  • *The SolenoidA long wire wound in the form of a helical coil is known as a solenoid.

  • Also, the magnetic field dB due to the current dI in dy can be found as,From the figure, the current for a length increment dy is

    1 (y2+R2) 1/2

    y2 (y2+R2) 3/2

    R2 (y2+R2) 3/2

  • *So, dB = ( ) cos d => B = ( ) cos d = ( ) (sin 2 sin 1 )

    B = (sin 2 sin 1 ) j

    where n = , number of turns per unit lengthThis formula represents the magnetic field along the centroid axis of a finite solenoid.For infinite long solenoid, it is assumed that 1 = -/2 and 2 = /2.B = 0 n I j

    0 I N 2 L

    0 I N 2 L

    21

    0 I N 2 L

    0 n I 2

    NL

  • *Example 17.3 : Since the length of the solenoid is quite large in comparative with its diameter, the magnetic field near its middle is approximately uniform. It is therefore reasonable to consider it as a case of infinite solenoidB = 0 n I j.Solution A solenoid has 300 turns wound around a cylinder of diameter 1.20 cm and length 14.0 cm. If the current through the coils is 0.410 A, what is the magnitude of the magnetic field inside and near the middle of the solenoid.The number of turns per unit length (n) isn = N/L = (300 turns) / (0.14 m) = 2.14 103 turns / mTherefore, the magnetic field inside and near the middle of the solenoid is,B = 0 n I j = (4 10-7 Tm/A) (2.14 103 turns / m) (0.410 A) = 1.10 10-3 T

  • *Magnetic fields of a finite solenoid