Dipole in Uniform Magnetic field
Most people are familiar with magnets primarily as toys or as decorative items on a refrigerator door
Magnetism has a much broader range of applications
Considering two opposite poles (i.e., North and South poles) as a system is known as a magnetic dipole
Therefore, based on the definition we can say that a bar magnet forms a magnetic dipole
Let us see what happens when we place a Dipole in Magnetic field
When a bar magnet or magnetic dipole is placed in a uniform Magnetic field, each pole will experience a force
Due to this Force, the Dipole is bound to experience a Torque
Let us see the Force and Torque acting on a Magnetic Pole in a field
When an electric charge is placed in an Electric field, Force acts on the charge due to the Electric Field Intensity
Similarly, a Magnetic pole experiences a Force due to a Magnetic Field
The Force acting on the Magnetic Pole is given by
$F→=mB→$
Where,
$m$
=The Magnetic pole strength
$B→$
=Magnetic induction
The pole strength acting on the North Pole is
$+m$
, so the Force acting on it will be
$F→=mB→$
The direction of the Force acting on the North magnetic Pole will be in the direction of the Magnetic Field
The pole strength of the South magnetic Pole will be
$−m$
, so the Force acting on it will be
$F→=−mB→$
The direction of the Force acting on the South magnetic Pole will be opposite to the direction of Magnetic Field
Now, let us consider a Dipole placed in a Uniform Magnetic Field
$B→$
at an angle
$θ$
The Force acting on the dipole is zero since the forces acting on the individual poles are equal and opposite
The Torque acting on a dipole can be mathematically expressed as,
$τ=F×$
perpendicular distance between North Pole N and South Pole P
The vertical distance between the North Pole and the South Pole is
$NP$
In
$△SPN$
,
$sinθ=SNNP ∴NP=SNsinθ=2lsinθ$
Replacing the value of
$NP=2lsinθ$
,
$τ=mB×2lsinθ$
Rearranging the terms,
$τ=(m)(2l)Bsinθ∴τ=MBsinθ$
Where,
$M=m(2l)$
is the Magnetic Moment on the Dipole
The equation of Torque can be written in vector form as,
$τ→=M→×B→$
In the equation
$τ=MBsinθ$
, if
$θ=0_{o}$
or
$θ=180_{o}$
, the Dipole is lying Parallel or Anti-Parallel to the Magnetic Field
In this case, the Torque acting on the Dipole will be Zero
This is because the Torques acting on the North Pole
$mB$
and South Pole
$−mB$
will be balanced
Now, if
$θ=90_{o}$
,
$sin90_{o}=1$
In this case, the Torque acting on the Dipole will be
$τ=MB$
Additionally, if
$B=1T$
, the Torque acting on the Dipole will be equal to Magnetic Moment
$M$
The Torque acting on a Dipole placed in a Unit Magnetic field perpendicular to the direction of Magnetic field is known as the Magnetic Dipole Moment
Let us calculate the Work done on the Dipole
Work is done whenever a Torque is used to rotate a Dipole by an angle
Work done can be mathematically expressed as
Work done = Torque x Angular Displacement in the direction of the Force
The Work done to rotate a bar magnet or Dipole by a small angle
$θ$
is given by,
$dW=τdθ$
Now, further Work done is required to move the Dipole from an initial angle
$θ_{0}$
to final angle
$θ$
This Work done is calculated by,
$W=∫dW$
$∫dW=∫_{θ_{0}}MBsinθdθ⇒W=MB∫_{θ_{0}}sinθdθ$
$⇒W=MBθ_{0}_{θ}⇒W=MB[cosθ_{0}−cosθ]$
If
$θ_{0}=0$
,
$W=MB[1−cosθ]$
If
$θ_{0}=90_{o}$
,
$W=−MBcosθ∴W=−M→.B→$
Since,
$A→.B→=ABcosθ_{0}$
This Work done will be stored in the form of Potential Energy. Therefore,
$P.E=−MBcosθ$
Revision
A magnet in which opposite poles (i.e., North and South poles) are on opposite sides of the magnet is called a Dipole magnet
When a bar magnet or magnetic dipole is placed in a uniform Magnetic field, it will experience a torque
The Force acting on a Dipole multiplied by the perpendicular distance between the Poles is the Torque acting on the Dipole
The work done on the dipole if
$θ_{0}=90_{o}$
, will be
$W=−M→.B→$
The End