Magnetic Force and Torque { The Generator }

(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=neAv_d
  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(v_d X B)
  where e is the amount of charge on an electron
  
• Total number of electron in length dl of the wire
  N=nAdl
  
• Thus magnetic force on wire of length dl is
  dF=(nAdl)(ev_d X B)
  if we denote length dl along the direction of current by the vector dl the above equation becomes
  dF=(nAev_d)(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
  F_AB=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
  F_CD=ihB
  and direction of F_CD is normal to the page but in the downwards direction going into the page
  
• The forces F_AB and F_CD are equal in magnitude and opposite in direction and hence they constitute a couple
  
• Torque τexerted by this couple on rectangular loop is
  τ=IhlB
  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
  τ=IhlB
  =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
  
  
  
  
  
  τ=NIABsinθ
  
• 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 ,
  τ=(NIA)Bsinθ
  if we take NIA as magnetic dipole moment (m) analogus to electric dipole moment (p),we have
  m=NIA                   -- (18)
  then
  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