The magnetic field B generated by wire 1 at distance r (where wire 2 is located) is
where I1 is the current in wire 1. So, the force exerted on a segment
of wire 2 is
By susbstituting B inside the expression of F, we find the formula for the force per unit of length between the two wires:
where r is the distance between the two wires.
By susbstituting the numbers of the problem, we get
Answer:
(a) 110 rev/ min
(b) 5/6
Explanation:
As per the conservation of linear momentum,
L ( initial ) = L ( final )
I' ω' = ( I' + I'' ) ωf
I' is the rotational inertia of first wheel and I'' is the rotational inertia of second wheel which is at rest.
(a)
So, ωf = I' ω' / ( I' + I'' )
As I'' = 5I'
ωf = I' ω' / ( I' + 5I' )
ωf = ω'/ 6
now we know ω' = 660 rev / min
therefore ωf = 660/6
= 110 rev/ min
(b)
Initial kinetic energy will be K'
K' = I'ω'² / 2
and final K.E. will be K'' = ( I' + I'' )ωf² / 2
K'' = ( I' + 5I' ) (ω'/ 6)²/ 2
K'' = 6I' ω'²/72
K'' = I' ω'²/ 12
therefore the fraction lost is
ΔK/K' = ( K' - K'' ) / K'
= {( I'ω'² / 2) - (I' ω'²/ 12)} / ( I'ω'² / 2)
= 5/6
I believe the answer is:
~D. Very dusty irregular galaxy.
Hope this helps!!!