Magnetic Dipole

Magnetics of Class 12

A small current carrying loop acts like a magnetic dipole. The magnitude of a magnetic dipole moment or magnetic moment Magnetic Dipole. It is defined as the product of the current in a flat current - carrying loop and the area enclosed by it.

Thus, Magnetic Dipole = IMagnetic Dipole

The direction of magnetic moment coincides with the direction of the area vector (which is the direction of the magnetic field).

If the loop contains N number of turns, the magnetic moment is given by

Magnetic Dipole= NIMagnetic Dipole (3.19)

 

Magnetic Dipole

Example: 3.12

A circular loop of wire of radius R carrying a current I is bent about its diameter along two mutually perpendicular planes as shown in the figure (3.20).

Find the magnetic moment of the loop.

Solution

The magnetic moment of the semicircle in the yz-plane is along the x-axis and that in the xz-plane is along the y-axis.

Thus, Magnetic Dipole = Magnetic Dipole

Magnetic Dipole= Magnetic Dipole

Magnetic Dipole

 

The total magnetic moment is given by

Magnetic Dipole = pxi+ py j

or Magnetic Dipole = Magnetic Dipole A.m2

Magnetic Dipole

Example : 3.13

A circular ring of radius R carries a charge q uniformly distributed on its circumference. It is rotated at constant angular speed ω about the centroidal axis perpendicular to its plane. Find its magnetic moment.

Magnetic Dipole

Solution

The motion of charges along with the ring produces current I which is given by

I = Magnetic Dipole

Thus, pm = IA = Magnetic Dipole (πR2) = 1/2 ωqR2

 

Example : 3.14

A rotating rod is pivoted about one end and carries a total charge q uniformly distributed along its length L (fig. 3. 22 a). If the rod rotates with angular velocity ω, compute its magnetic moment μ.

Magnetic Dipole

Solution

We visualize the rotating rod of charge as a series of concentric current loop, each carrying a current dI. The charge per unit length λ along the rod is

λ = q/L

Now dq = λdl and dI = Magnetic Dipoleλdl

The magnetic moment of this differential current loop is given by

dpm = dI (πl2) = Magnetic Dipoleπl2 = Magnetic Dipolel2 dl

On integrating ,

Pm = Magnetic Dipole

Substituting for λ, we have

Pm = Magnetic Dipole

 

1.

Electric Dipole

Magnetic Dipole

Magnetic Dipole

Magnetic Dipole

2.

Electric Field along the axis

Magnetic Dipole|| = Magnetic Dipole

Magnetic Field along the axis

Magnetic Dipole|| = Magnetic Dipole

3.

Torque

Magnetic Dipole= Magnetic DipoleE × Magnetic Dipole

Torque

Magnetic Dipole= Magnetic Dipolem × Magnetic Dipole (3.20)

4.

Potential Energy

U = − Magnetic DipoleE .Magnetic Dipole

Potential Energy

U = − Magnetic Dipolem.Magnetic Dipole (3.21)

 

Example: 3.15

The arrangement is same as shown in Fig. (3.23a ).

(a) Find the potential energy of the loop

(b) Find the work done, to increase the spacing between the wire and the loop from a to 2a.

Solution

(a) Since magnetic field is non-uniform over the loop therefore we cannot straight away use equation (3.21) to find the potential energy of the loop.

A small loop of length l and width dx is assumed as shown in fig, (3.23 b). The magnetic moment of this loop is given by

dpm = I2ldx

Magnetic Dipole

The direction of the magnetic moment is perpendicular to the plane of paper pointing inwards. Thus, the potential energy of this loop is given by

dU =-d Magnetic Dipolem.Magnetic Dipole= −dpmB

where B is the magnetic field at the position of this element

i.e. B = Magnetic Dipole

Thus, dU = Magnetic Dipole

The total potential energy is

U = Magnetic Dipole

U = Magnetic Dipole

Magnetic Dipole

(b) The initial and final potential energies of the loop are

Ui = Magnetic Dipole

Ur = Magnetic Dipole

The work done is given by

W = Ur − Ui = Magnetic Dipole

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