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Torque on a current carrying rectangular loop in a magnetic field|Magnetism




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 parallel to the plane of the coil as shown below in the figure



    Torque on a current carrying rectangular loop in a magnetic field




  • 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
    FCD=ihB
    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
    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
  • Similarly 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



    Magnetic force on the rectangular coil

    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) analogous 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 considering a rectangular loop but these equations are valid for plane loops of any shape







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