Chapter 4: Magnetic Field - YSL Chapter 4 CP3 1 FYSL Chapter 4: Magnetic Field 4.1 Magnetic Field L.O

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  • Chapter 4

    CP3 1 FYSL

    Chapter 4: Magnetic Field

    4.1 Magnetic Field

    L.O 4.1.1 Define magnetic field

     Magnetic field is defined as the region around a magnet where a magnetic force can be

    experienced.

     Magnetic field has two poles, called north (N) and south (S). These magnetic poles are

    always found in pairs whereas a single magnetic pole has never been found.

     Like poles (N-N or S-S) repel each other.

     Opposite poles (N-S) attract each other.

    L.O 4.1.2 Identify magnetic field sources and sketch their magnetic field lines

     Magnetic field lines are used to represent a magnetic field.

     The characteristics of magnetic field lines:

     The lines do not intersect one another

     The lines form a closed loop: magnetic field lines leave the North-pole and enter the

    South-pole.

     The lines are closer together at the poles. (The number of lines per unit cross-

    sectional area is proportional to the magnitude of the magnetic field.)

     Two sets of magnetic field lines can be superimposed to form a resultant magnetic

    field line.

     Magnetic field can be represented by crosses or by dotted circles as shown in figures

    below:

    Magnetic field lines enter

    the page perpendicularly

    Magnetic field lines leave

    the page perpendicularly

  • Chapter 4

    CP3 2 FYSL

     The pattern of the magnetic field lines can be determined by using two methods:

     Using compass needles

     Using sprinkling iron filings on paper

     Magnetic field sources:

    i. Bar magnet

    One bar magnet Horseshoe or U magnet

    Two bar magnets (unlike pole)

    Two bar magnets (like pole)

  • Chapter 4

    CP3 3 FYSL

    ii. Current-carrying conductor

     A stationary electric charge is surrounded by an electric field only.

     When an electric charge moves, it is surrounded by an electric field and a

    magnetic field. The motion of the electric charge produces the magnetic field.

    A circular coil A long straightwire A solenoid

    iii. Earth magnetic field

    Example

    Note that the Earth’s “North

    Pole” is really a south magnetic

    pole, as the north ends of

    magnets are attracted to it.

    N S

    N S

    N S

    S N

  • Chapter 4

    CP3 4 FYSL

    4.2 Magnetic Field Produced by Current-Carrying Conductor

    L.O 4.2.1 Use magnetic field equations

    a) Long Straight Wire

    View (a) from the top:

    Magnitude of the magnetic field at any point

    from the conductor (wire):

    where

    B = magnetic field strength / magnetic flux

    density (T)

    I = current in the wire (A)

    r = perpendicularly distance of P from

    the wire (m)

    μo = constant of proportionality

    known as the permeability of

    free space (vacuum)

    = 4π x 10 -7

    Henry per metre (H m -1

    )

    The direction of magnetic field around the

    wire or coil can be determined by using the

    right hand grip rule as shown in Figure(b).

     Thumb – direction of current

     Other fingers – direction of magnetic field (clockwise OR anticlockwise)

    r

    I B

    2

    0

    r B •

  • Chapter 4

    CP3 5 FYSL

    b) Circular Coil

    View from the top:

    (a)

    (b)

    Magnitude of the magnetic field at the centre

    of the circular coil:

    For ONE circular coil

    For N circular coils

    where

    R = radius of the circular coil.

    µ0 = permeability of free space

    4π × 10 -7

    H m -1

    I = current

    N = number of coils (loops)

    The direction of magnetic field around the

    wire or coil can be determined by using the

    right hand grip rule as shown in Figure(a).

    Example of multiple circular loops:

    B

    B

  • Chapter 4

    CP3 6 FYSL

    c) Solenoid

    View from the top:

    (a)

    The magnitude of magnetic field intensity at

    the centre (mid-point/ inside) of N turn

    solenoid is given by

    Since , therefore it can be written as

    The magnitude of magnetic field intensity at

    the end of N turn solenoid is given by:

    where

    n = numbers of turns per unit length

    µ0 = permeability of free space

    4π × 10 -7

    H m -1

    I = current

    N = number of coils (loops)

    The directions of the fields can be found by

    viewing the current flows in the solenoid from

    both end or applying the right hand grip rule

    as shown in Figure (a).

     Thumb – north pole

     Other fingers – direction of current in solenoid.

  • Chapter 4

    CP3 7 FYSL

    Example

    Question Solution

    Determine the magnetic field strength at

    point X and Y from a long, straight wire

    carrying a current of 5 A as shown below.

    A circular coil having 400 turns of wire in air

    has a radius of 6 cm and is in the plane of the

    paper. What is the value of current must

    exist in the coil to produce a flux density of

    2 mT at its center?

    An air-core solenoid with 2000 loops is 60

    cm long and has a diameter of 2.0 cm. If a

    current of 5.0 A is sent through it, what will

    be the flux density within it?

    A solenoid is constructed by winding 400

    turns of wire on a 20 cm iron core. The

    relative permeability of the iron is 13000.

    What current is required to produce a

    magnetic induction of 0.5 T in the center of

    the solenoid?

  • Chapter 4

    CP3 8 FYSL

    Question Solution

    Two straight parallel wires are 30 cm apart

    and each carries a current of 20 A. Find the

    magnitude and direction of the magnetic field

    at a point in the plane of the wires that is 10

    cm from one wire and 20 cm from the other

    if the currents are

    i. in the same direction, ii. in the opposite direction.

    Exercise

    Question

    Two long straight wires are oriented perpendicular to the page as shown in Figure below.

    The current in one wire is I1 = 3.0 A pointing into the page and the current in the other wire is

    I2 = 4.0 A pointing out of the page. Determine the magnitude and direction of the nett

    magnetic field intensity at point P. (Given 0=4  x 10 -7

    H m -1

    )

    Answer: 8.93×10 -6

    T, 63.1° below +ve x-axis

    A 2000 turns solenoid of length 40 cm and resistance 16  is connected to a 20 V supply.

    Find the magnetic flux density at the end of the axis of the solenoid.

    Answer: 3.95×10 -3

    T

  • Chapter 4

    CP3 9 FYSL

    4.3 Force on a Moving Charged Particle in a Uniform Magnetic Field

    L.O 4.3.1 Use magnetic force equation

     A stationary electric charge in a magnetic field will not experience any force. But if the

    charge is moving with a velocity, v in a magnetic field, B then it will experience a

    force. This force known as magnetic force.

     The magnitude of the magnetic force can be calculated by using the equation below :

     

      

     

    

    BvqF

    sinqvBF 

     The direction of the magnetic force can be determined by usingright hand rule:

    Example

    Determine the direction of the magnetic force, exerted on a charge in each problems below:

    where

    q : magnitude of the charge

    : angle between ⃗ and ⃗⃗

    F

    v

    B

    IMPORTANT!

     Thumb indicates the direction

    of the magnetic force exerted

    on a positive charge.

     If the charge is negative,

    direction of the force is

    opposite.

  • Chapter 4

    CP3 10 FYSL

    L.O 4.3.2 Describe circular motion of a charge in a uniform magnetic field

    L.O 4.3.3 Use relationship FB = FC

     Consider a charged particle moving in a uniform magnetic field with its velocity

    perpendicular to the magnetic field.

     As the particle enters the region, it will experience a magnetic force which the force is

    perpendicular to the velocity of the particle. Hence the direction of its velocity changes

    but the magnetic force remains perpendicular to the velocity.

     Since the path is circle therefore the magnetic force FB contributes the centripetal force

    Fc (net force) in this motion. Thus

    CB FF 

    r

    mv Bqv

    2

    sin  and θ = 90°

    Bq

    mv r 

     The period of the circular motion, T makes by the particle is given by

    v

    r T

    2 

    Bq

    m T

    2 

     The frequency of the circular motion makes by the particle is given by

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