Magnetics 4 Question 2

3. An insulating thin rod of length l has a linear charge density ρ(x)=ρ0xl on it. The rod is rotated about an axis passing through the origin (x=0) and perpendicular to the rod. If the rod makes n rotations per second, then the time averaged magnetic moment of the rod is

(2019 Main, 10 Jan I)

(a) nρl3

(b) πnρl3

(c) π3nρl3

(d) π4nρl3

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Answer:

Correct Answer: 3. (d)

Solution:

(R2= radius of coil  having N loops) 

Key Idea A rotating charge constitutes a current. Hence, a rotating charged rod behaves like a current carrying coil. If charge q rotates with a frequency n, then equivalent current is I=qn and magnetic moment associated with this current is M=IA

where, A =area of coil or area swept by rotating rod. Let dq be the charge on dx length of rod at a distance x from origin as shown in the figure below.

The magnetic moment dm of this portion dx is given as

dm=(dI)Adm=ndqA[I=qn,dI=ndq]=nρdxA

where, ρ= charge density of rod=ρ0xl.

So, dm=nρ0xdxπx2l=πnρ0lx3dx

Total magnetic moment associated with rotating rod is sum of all the magnetic moments of such differentiable elements of rod.

So, magnetic moment associated with complete rod is

M=x=0x=ldm=0lπnρ0lx3dx=πnρ0l0lx3dx=πnρ0lx44=πnρ0l34

 at x=lρ=ρ0M=π4nρl3



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