Magnetic Moment of A Rotating Charged Rod

Magnetic Moment of A Rotating Charged Rod

Magnetic Moment:

In general for any closed loop magnetic moment is given by formula `\vec{μ}= ni \vec{A}` , where ` \vec{A}` represents the area vector of the loop determined as usual by right hand thumb rule.

Now, the primary problem for us is like this:

Prob. A rod with charge density ` ρ = ρ_{0}\frac{x}{L} ` is rotated about its one end along an axis perpendicular to its length. Then find the magnetic moment of the setup.

For a Revolving Charged Particle:

Before going into the problem, it is important to realise that we have been given a charged rod, and not a simple wire loop wherein a current flows. However as the rod shall rotate, the charges will tranverse through space thus creating an apparent current.

In any circular path, when a charged particle say q revolves it appears to create and average current, 

`<i> = qf = \frac{q}{T} = \frac{qω}{2π}`

Thus in that case, 

` μ = (\frac{qw}{2π})(π r^{2}) = \frac{qωr^{2}}{2}`

For the Rotating Rod:

We now follow a similar process for the rotating rod.

At a distance `x` from the axis consider an elemental charge element `dq = ρdx = ρ_{0} \frac{x}{L}dx `. Thence the magnetic moment due to this rotating `dq` is given by,

`dμ = dq f = dq (\frac{ω}{2π}) (π x^{2})`

` dμ = ρ_{0} \frac{x}{L} dx \frac{ω}{2} x^{2}`

` dμ =\frac {ρ_{0}ω}{2L} x^{3} dx `

Therefore for the whole rod we have, 

`  μ = \frac {ρ_{0}ω}{2L}  \int_{0}^{L}x^{3}dx `

`  μ =  \frac {ρ_{0}ω}{2L} • \frac{L^{4}}{4} =  \frac {ρ_{0}ω}{8}L^{3} `

Thus the magnetic moment of the setup is, 

` μ =  \frac {ρ_{0}ω}{8}L^{3} `

The above approach can be easily extended to similar problems.