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Grade: 10
        
A non conducting disc of radius R, charge Q is rotating about an axis passing through its centre and perpendicular to its plane,  with an angular velocity  'ω' . 
The magnetic moment of the disc is =?
a) 1/4 qωR^2    b)  1/2 qωR^2.   c) qωR.     d) 1/2qωR^2
 
 
9 months ago

Answers : (2)

Arun
23864 Points
							
surface charge density σ = q/πr²
Now let us take a small cross section of length dR whose radius varies from R to R + dR
The charge on this annular region is ,
dq = 2πRσdR 
= 2πR x q/πr² x dR
= 2q/r² x RdR
This charge dq passes through a line once in a time, T = 2π/ω
So charge crossing per unit time i.e. current dI = dq/T
= dqω/2π
=2q/r² x RdR x ω/2π
The area enclosed by this current = πR²
Hence, magnetic moment due to current loop ,
dM = πR² dI
=πR² x  2q/r² x RdR x ω/2π
= ωq/r²  x R³dR
Taking integration on both the sides where limit of R varies from 0 to R we get,
M = ωq/r²  x 
=> M = ωq/r²  x r⁴/4
=> M = ωqr²/4
Hence the moment of inertia will be ωqr²/4
 
9 months ago
Khimraj
3008 Points
							

dM = πR² dI

=πR² x  2q/r² x RdR x ω/2π

= ωq/r²  x R³dR

M = ωq/r²  x \int\limits^r_0 {R^3} \, dR \\

=> M = ωq/r²  x r⁴/4

=> M = ωqr²/4

......................................................................

7 months ago
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