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Question: A uniform rod of length \(L\) and large \(q\) is rotated with a constant angular velocity \(\omega \...

A uniform rod of length LL and large qq is rotated with a constant angular velocity ω\omega about an axis passing through its end and perpendicular to its length. Find the magnetic moment associated with its rotation.
A) 112qL2ω\dfrac{1}{{12}}q{L^2}\omega
B) 124qL2ω\dfrac{1}{{24}}q{L^2}\omega
C) 13qL2ω\dfrac{1}{3}q{L^2}\omega
D) 16qL2ω\dfrac{1}{6}q{L^2}\omega

Explanation

Solution

When the uniform rod carrying a charge rotates with a constant angular velocity, a current gets generated in the rod. The magnetic moment of the rotation depends on the current generated for the entire rod and the area of the circle described by the rod as it rotates.

Formulas used:
-The magnetic moment of a rotating rod is given by, M=0rdMM = \int\limits_0^r {dM} where dMdM is the magnetic moment due to a small element of the rod; dM=dIAdM = dI \cdot A where dIdI is the current through the small element and AA is the area of the circle described by it.
-The current generated by the rotation of the small element of charge dqdq of a rod is given by, dI=dqTdI = \dfrac{{dq}}{T} where TT is the period of rotation.
-The period of rotation of a rod is given by, T=2πωT = \dfrac{{2\pi }}{\omega } where ω\omega is the constant angular velocity of the rotation.

Complete step by step answer.
Step 1: Sketch a rough figure of the rod to find the current through a small element dxdx.

In the above figure, we consider a small element of length dxdx at a distance xx from its axis of rotation. The length of the rod is LL and it rotates with a constant angular velocity ω\omega .
The small element carries a charge dqdq given by, dq=qLdxdq = \dfrac{q}{L}dx
The current generated by the rotation of the small element will be dI=dqTdI = \dfrac{{dq}}{T} ------ (1) where TT is the period of rotation.
Substituting for dq=qLdxdq = \dfrac{q}{L}dx and T=2πωT = \dfrac{{2\pi }}{\omega } in equation (1) we get, dI=(qLdx)(2πω)dI = \dfrac{{\left( {\dfrac{q}{L}dx} \right)}}{{\left( {\dfrac{{2\pi }}{\omega }} \right)}}
Now on simplifying, it becomes dI=(qω2πLdx)dI = \left( {\dfrac{{q\omega }}{{2\pi L}}dx} \right)
Thus the current through the small element dxdx is dI=(qω2πLdx)dI = \left( {\dfrac{{q\omega }}{{2\pi L}}dx} \right) .
Step 2: Express the relation for the magnetic moment of the small element dxdx.
The magnetic moment of a rotating small element of the rod is given by, dM=dIAdM = dI \cdot A -------(2)
where dIdI is the current generated due to the charge in the small element dxdx and AA is the area of the circle described by the small element as it rotates.
Here, the small element dxdx describes a circle of radius xx and thus the area of the circle will be A=πx2A = \pi {x^2}
Now we substitute for A=πx2A = \pi {x^2} and dI=(qω2πLdx)dI = \left( {\dfrac{{q\omega }}{{2\pi L}}dx} \right) in equation (2) to get, dM=(qω2πLdx)×πx2dM = \left( {\dfrac{{q\omega }}{{2\pi L}}dx} \right) \times \pi {x^2} or on simplifying we get, dM=qω2Lx2dxdM = \dfrac{{q\omega }}{{2L}}{x^2}dx -------- (3).
Step 3: Using equation (3) the magnetic moment of the entire rod is found out.
Equation (3) gives the magnetic moment of the small element dxdx as dM=qω2Lx2dxdM = \dfrac{{q\omega }}{{2L}}{x^2}dx .
Then to find the magnetic moment MM for the entire rod we integrate equation (3) for the length of the rod.
i.e., M=0LdMM = \int\limits_0^L {dM} ------- (4).
Substituting equation (3) in (4) we obtain M=0Lqω2Lx2dxM = \int\limits_0^L {\dfrac{{q\omega }}{{2L}}{x^2}dx}
On evaluating the integral we get, M=qω2L[x33]0LM = \dfrac{{q\omega }}{{2L}}\left[ {\dfrac{{{x^3}}}{3}} \right]_0^L
Then on applying the limit we get, M=qω2L[L33]=qωL26M = \dfrac{{q\omega }}{{2L}}\left[ {\dfrac{{{L^3}}}{3}} \right] = \dfrac{{q\omega {L^2}}}{6}
\therefore the magnetic moment of the rod is M=16qL2ωM = \dfrac{1}{6}q{L^2}\omega .

Hence the correct option is D.

Note: The limits of integration for evaluating the magnetic moment of the entire rod is given varies from 00 to LL. This is because the radius of the circle described by the entire rod will be LL as the axis of rotation passes through one of the ends of the uniform rod. The limits of integration vary depending on the position of the rotational axis.