Physics 12th Magnetic Field

Magnetic Field

(1) Introduction

  • We have allready studied about thermal effects of current and now in the present chapter we are studied about magnetic effect of current.
  • Earlier it was believe that there is no connection between electric and magnetic force and both of them are completely different.
  • But in 1820 Oersted showed that the electric current through a wire deflect the magnetic needle placed near the wire and the direction of deflection of needle is reversed if we reverse the direction of current in the wire.
  • So, Oersted's experiments establishes that a magnetic field is assoiated with current carrying wire.
  • Again if we a magnetic needlle near a bar magnet it gets deflectid and rests in some other direction.
  • This needle experiences the tourque which turn the needle to a definite direction.
  • Thus, the reagion near the bar magnet or current carrying where magnetic needle experience and suffer deflection is called magnetic field.

(2) The Magnetic Field

  • We all ready know that a stationery charges gets up a electric field E in the space surrounding it and this electric field exerts a force F=q0E on the test charge q0 placed in magnetic field.
  • Similarly we can describe the intraction of moving charges that, a moving charge excert a magnetic field in the space surrounding it and this magnetic field exert a force on the moving charge.
  • Like electric field, magntic field is also a vector quantity and is represented by symbol B
  • Like electric field force which depend on the magnitude of charge and electric field, magnetic force is propotional to the magnitude of charge and the strength of magnetic field.
  • Apart from its dependence on magnitude of charge and magnetic field strength magnetic force also depends on velocity of the particle.
  • The magnitude magnetic force increase with increase in speed of charged particle.
  • Direction of magnetic force depends on direction of magnetc field B and velocity v of the chared particle.
  • The direction of magnetic force is not alonge the direction of magnetic field but direction of force is always perpendicular to direction of both magnetic field B and velocity v
  • Test charge of magnitude q0 is moving with velocity v through a point P in magnetic field B experience a deflecting force F defined by a equation
    F=qv X B 
  • As mentioned earlier this force on charged particle is perpendicular to the plane formed by v and B and its direction is determined right hand thumb rule.

  • When moving charge is positive the direction of force F is the direction of advance of hand screw whose axis is perpendicular to the plane formed by v and B.

  • Direction of force would be opposit to the direction of advance screw for negative charge moving in same direction.
  • Magnitude of force on charged particle is
    where θ is the angle between v and B.
  • If v and B are at right angle to each other i.e. θ=90 then force acting on the particle would be maximum and is given by
    Fmax=q0vB                   ----(3)
  • When θ=180 or θ=0 i.e. v is parallel or antiparallel to B then froce acting on the particle would be zero.
  • Again from equation 2 if the velocity of the palticle in the magnetic field is zero i.e., particle is stationery in magnetic field then it does not experience any force.
  • SI unit of strength of magnetic field is tesla (T). It can be defined as follows
    for F=1N,q=1C and v=1m/s and θ=90
    Thus if a charge of 1C when moving with velocity of 1m/s along the direction perpendicular to the magnetic field experiences a force of 1N then magnitude of field at that point is equal to 1 tesla (1T).
  • Another SI unit of magnetic field is weber/m2 Thus
    1 Wb-m-2=1T=1NA-1m-1
    In CGS system, the magnetic field is expressed in 'gauss'. And 1T= 104 gauss. Dimention formula of magnetic field (B) is [MT-2A-1]

(3) Lorentz Force
  • We know that force acting on any charge of magnitude q moving with velocity v inside the magnetic field B is given by
    F=q(v X B)
    and this is the magnetic force on charge q due to its motion inside magnetic field.
  • If both electric field E and magnetic field B are present i.e., when a charged particle moves through a reagion of space where both electric field and magnetic field are present both field exert a force on the particle and the total force on the particle is equal to the vector sum of the electric field and magnetic field force.
    F=qE+q(v X B)                   (4)
  • This force in equation(4) is known as Lorentz Force.
  • Where important point to note is that magnetic field is not doing any work on the charged particle as it always act in perpendicular direction to te motion of the charge.
(4)Motion of Charged Particle in The Magnetic Field
  • As we have mentioned earlier magnetic force F=(vXB) does not do any work on the particle as it is perpendicular to the velocity.
  • Hence magnetic force does not cause any change in kinetic energy or speed of the particle.
  • Let us consider there is a uniform magnetic field B perpendicular to the plane of paper and directed in downward direction and is indicated by the symbol C in figure shown below.
  • Now a charge particle +q is projected with a velocity v to the magnetic field at point O with velocity v directed perpendicular to the magnetic field.
  • Magnetic force acting on the particle is
    F=q(v X B) = qvBsinθ
    Since v is perpendicular to B i.e., angle between v and B is θ=90 Thus charged particle at point O is acted upon by the force of magnitude
    and the direction of force would be perpendicular to both v and B
  • Since the force f is perpendicular to the velocity, it would not change the magnitude of the velocity and the peffect of this force is only to change the direction of the velocity.
  • Thus under the action of the magnetic force of the particle will more along the circle perpendicular to the field.
  • Therefore the charged particle describe an anticlockwise circular path with constant speed v and here magnetic force work as centripetal force. Thus
    where radius of the circular path traversed by the particle in the magnetic in field B is given as
    r=mv/qB                   ---(5)
    thus radius of the path is proportional to the momentum mv pof the charged particle.
  • 2πr is the distance traveled by the particle in one revolution and the period T of the complete revolution is
    T=2 πr /v
    From equation(5)
    time period T is
    T=2πm/qB                   (6)
    and the frequency of the particle is f=1/T=qB/2πm                   (7)
  • From equation (6) and (7) we see that both time period and frequency does not dependent on the velocity of the moving charged particle.
  • Increasing the speed of the charged particle would result in the increace in the radius of the circle. So that time taken to complete one revolution would remains same.
  • If the moving charged particle exerts the magnetic field in such a that velocity v of particle makes an angle θ with the magnetic field then we can resolve the velocity in two components
    vparallel : Compenents of the velocity parallel to field
    vperpendicular :component of velocity perpendicular to magnetic field B 
  • The component vpar would remain unchanged as magnetic force is perpendicular to it.
  • In the plane perpendicular to the field the particle travels in a helical path. Radius of the circular path of the helex is
    r=mvperpendicular/qB=mvsinθ/qB                   (8)

(5) Cyclotron

  • Cyclotron is a machine for producing high energy particles ,first developed by E.O.Lawrence and M.S.Livingston in 1931.Figure below shows the path of a charged particle in a cyclotron


  • cyclotron consists of two horizontal D-shaped hollow metal segments D1 and D1 with a small gap between them
  • These D'ees are placed in between the poles of a large electromagnet so that that magnetic field is Perpendicular to the plane of the D'ees
  • The whole space inside the D'ees is evacuated to pressure of about 10-6 mm of Hg
  • An ion source S is kept at the center between the D'ees
  • The two D'ees are connected to the terminals of high freqency oscillating A.C circuit.This changes the charge of each D'ees several million time per sec
Theory and working

  • Suppose that any instant ,alternating potential is in the direction which makes D1 positive and D2 negative
  • A positive ion starting from source S will be attracted by the Dee D2
  • Since Uniform magnetic field B acts at right angles to the plane of the Dees ,the positively charged ion of the charge q and mass m will move in a circular motion of radius
    where v is the speed of the particle and it is constant
  • After traversing half a cycle the ion comes to the edge of D2.If we adjust the frequency of the oscillator in such a way that by the time ,ion comes to the edge of D2,potential difference changes direction so as to make D1 negative and D2positive.
  • The ion will then get then attracted to D1 and its speed will increasae due to acceleration
  • Once inside D1 ,the ion is now in electric field free zone and again it will move in a circular path with constant speed which is higher then the previous constant speed in D2.Radius of the path in D1 will be larger then D2
  • After traversing the semi -circular path in D1 ,the ion will come to the egde of D1 where if the direction of electric field changes ,it will receive additional energy
  • This way the ion will continue travelleing in semi circles of increasing radii every time it goes from D2 to D1 and from D1to D2 
  • Time taken by the ion to traverse the semi-circular path in the Dee is given by
  • Thus by adjusting the magnetic field B,t can be made the same as that required to change the potential of the D1 and D2,so that positive charge ion always crosses the alternating electric field across the gap in correct phase
  • Ions gain tremendous amount of energy after traversing through reversal rotation.when they come near the circumference of the Dees,an auxillary electric field is used to deflect them from the circular path to eventually reach a target
  • Frequency F of charged particle moving in a cyclotron is
    f=ω/2π =υ/2πr=Bυ/2πm                   -- (10)
    where υ=1/2t
  • If f and B are adjusted to keep charged ion always in phase each time ,the ion crosses the gap .it receive additional energy and at the same time it describes a flat spiral of increasing radius
  • KE of ion emerging from the cyclotron if R is radius of the D'e is

  • Above relation shows that the maximum energy attained by the ion is limited by the radius R of the Dees ,magnetic field B or the frequency f 
  • Maximum energy acquired by the charged particle in a particular cyclotron is independent of the alternating potential i.e when the voltage is small the ion makes a large number of the turns before reaching the periphery and for the large voltage number of turns is small.Total energy remians the same in both the cases,provided both B and R are unchanged
  • These days cyclotron are not in wide use but others based on principle of cylotron are used
(6) Magnetic force on a current carrying wire

  • We know that current flowing in a conductor is nothing but the drift of free electron's from lower potential end of the conductor to the higher potential end
  • when a current carrying conductor is placed in a magnetic field ,magnetic forces are exerted on the moving charges with in the conductor
  • Equation -1 which gives force on a moving charge in a magnetic field can also be used for calculating the magnetic force exerted by magnetic field on a current carrying conductor (or wire)
  • Let us consider a straight conducting wire carrying current I is placed in a magnetic field B(x).Consider a small element dl of the qire as shown below in the figure

  • Drift velocity of electrons in a conductor and current I flowing in the conductor is given by I=neAvd
    Where A is the area of cross-section of the wire and n is the number of free electrons per unit volume
  • Magnetic force experienced by each electron in presence of magnetic field is
    F=e(vd X B)
    where e is the amount of charge on an electron
  • Total number of electron in length dl of the wire
  • Thus magnetic force on wire of length dl is
    dF=(nAdl)(evd X B)
    if we denote length dl along the direction of current by the vector dl the above equation becomes
    dF=(nAevd)(dl X B)
    or dF=I(dl X B)                   -- (12)
    where the quantity IdL is known as current element
  • If a straight wire of length l carrying current I is placed in a uniform magnetic field then force on wire would be equal to
    dF=I(L X B)                   -- (13)

Direction of force
  • Direction of force is always perpendicular to the plane containing the current element IdL and magnetic field B

  • Direction of force when current element IdL and B are perpendicular to each other can also be find using either of the following rules
    i) Fleming'e left hand rule:-
    If fore finger ,the middle finger and thumb of the left hand are stretched in such a way that the all are mutually perpendicular to each other then,if the fore finger points in the direction of the field (B) and middle finger points in the direction of current I ,the thumb will point in the direction of the force
    ii) Right hand palm Rule:
    Stretch the finger and thumb of the right hand so that they are perpendicular to each other .If the fingers point in the direction of current I and the palm in the direction of field B then the thumb will point in the direction of force
(7) Torque on a current carrying rectangular loop in a magnetic field
  • Consider a rectangular loop ABCD being suspended in a uniform magnetic field B and direction of B is paralle to the plane of the coil as shown below in the figure

  • Magnitude of force on side AM according to the equation(13) is
    FAB=IhB ( angle between I and B is 900)
    And direction of force as calculated from the right hand palm rule would be normal to the paper in the upwards direction
  • Similarly magnitude of force on CD is
    and direction of FCD is normal to the page but in the downwards direction going into the page
  • The forces FAB and FCD are equal in magnitude and opposite in direction and hence they constitute a couple
  • Torque τexerted by this couple on rectangular loop is
    Since torque = one of the force * perpendicular distance between them
  • No force acts on the side BC since current element makes an angle θ=0 with B due to which the product (ILXB) becomes equal to zero
  • Similary on the side DA ,no magnetic force acts since current element makes an angle θ=1800 with B
  • Thus total torque on rectangular current loop is
    =IAB                   --(15)
    Where A=hl is the area of the loop
  • If the coil having N rectangular loop is placed in magnetic field then torque is given by
    τ=NIAB                   ----(16)
  • Again if the normal to the plane of coil makes an angle θ with the uniform magnetic field as shown below in the figure then

  • We know that when an electric dipole is placed in external electric field then torque experienced by the dipole is
    τ=P X E=PEsinθ
    Where P is the electric dipole moment 
  • comparing expression for torque experienced by electric dipole with the expression for torque on a current loop i.e ,
    if we take NIA as magnetic dipole moment (m) analogus to electric dipole moment (p),we have
    m=NIA                   -- (18)
    τ=m X B                   -- (19)
  • The coil thus behaves as a magnetic dipole
  • The direction of magnetic dipole moment lies along the axis of the loop
  • This torque tends to rotate the coil about its own axis .Its value changes with angle between the plane of the coil and the direction of the magnetic field
  • Unit of magnetic moment is Ampere.meter2 (Am2)
  • Equation (18) and (19) are obtained by comsidering a rectangular loop but thes equations are valid for plane loops of any shape

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