# Dear sir, Please explain the concept of magnetic moment in detail and there is a sum based on it please explain how to solve it. The sum is, An insulating rod of length l carries uniform charge q distributed on it. It is pivoted at middle point and rotated with frequency f about a fixed axis perpendicular to the rod and passing through the pivot so what is the magnetic moment of the system?

509 Points
12 years ago

consider a charge moving in a circular path ...let radius of its path of radius r ...

due to its motion it a current is produced along the path ...then motion this particle

can be compared to a charged ring of radius r, circulating with some angular velocity...

magnetic moment = M = IA = I(pir2)               .........................1

we have to calculate magnetic moment for rod , let mid point of rod is O ...

now  consider two small elements dq of thickness dx on the rod which are at distance x from center

on both sides...

this (dq+dq) charge is moving in circle so producing current ...

total charge= 2dq

charge per unit length = q/L

dq = (q/L)dx  so

total charge = (2q/L)dx                     .........................2

total current over 1 complete revolution (t = 2pi/w) = total charge/t = (wq/piL)dx    .......3            (w = angular frequency)

area of circle  = pix2 so

magnetic moment = (total current)(area)

=pix2(wq/piL)dx

dM = (wq/L)x2dx

integrating both sides

M = (wq/L) (x3/3)             lim 0 to L/2

M = (wq/L)(L2/24)

M = wqL2/24

now , w = 2pif so

M = piqL2/12

this is the required result