Why do planets rings surround them and not the sun
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Answer:
a) ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ ) N.m
b) ΔU = -0.000747871 J
c) w = 47.97 rad / s
Explanation:
Given:-
- The area of the circular ring, A = 4.45 cm^2
- The current carried by circular ring, I = 13.5 Amps
- The magnetic field strength, vec ( B ) = (1.05×10−2T).(12i^+3j^−4k^)
- The magnetic moment initial orientation, vec ( μi ) = μ.(−0.8i^+0.6j^)
- The magnetic moment final orientation, vec ( μf ) = -μ k^
- The inertia of ring, T = 6.50×10^−7 kg⋅m2
Solution:-
- First we will determine the magnitude of magnetic moment ( μ ) from the following relation:
μ = N*I*A
Where,
N: The number of turns
I : Current in coil
A: the cross sectional area of coil
- Use the given values and determine the magnitude ( μ ) for a single coil i.e ( N = 1 ):
μ = 1*( 13.5 ) * ( 4.45 / 100^2 )
μ = 0.0060075 A-m^2
- From definition the torque on the ring is the determined from cross product of the magnetic moment vec ( μ ) and magnetic field strength vec ( B ). The torque on the ring in initial position:
vec ( τi ) = vec ( μi ) x vec ( B )
= 0.0060075*( -0.8 i^ + 0.6 j^ ) x 0.0105*( 12 i^ + 3 j^ -4 k^ )
= ( -0.004806 i^ + 0.0036045 j^ ) x ( 0.126 i^ + 0.0315 j^ -0.042 k^ )
- Perform cross product:
- The initial torque ( τi ) is written as follows:
vec ( τi ) = ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ )
- The magnetic potential energy ( U ) is the dot product of magnetic moment vec ( μ ) and magnetic field strength vec ( B ):
- The initial potential energy stored in the circular ring ( Ui ) is:
Ui = - vec ( μi ) . vec ( B )
Ui =- ( -0.004806 i^ + 0.0036045 j^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )
Ui = -[( -0.004806*0.126 ) + ( 0.0036045*0.0315 ) + ( 0*-0.042 )]
Ui = - [(-0.000605556 + 0.00011)]
Ui = 0.000495556 J
- The final potential energy stored in the circular ring ( Uf ) is determined in the similar manner after the ring is rotated by 90 degrees with a new magnetic moment orientation ( μf ) :
Uf = - vec ( μf ) . vec ( B )
Uf = - ( -0.0060075 k^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )
Uf = - [( 0*0.126 ) + ( 0*0.0315 ) + ( -0.0060075*-0.042 ) ]
Uf = -0.000252315 J
- The decrease in magnetic potential energy of the ring is arithmetically determined:
ΔU = Uf - Ui
ΔU = -0.000252315 - 0.000495556
ΔU = -0.000747871 J
Answer: There was a decrease of ΔU = -0.000747871 J of potential energy stored in the ring.
- We will consider the system to be isolated from any fictitious forces and gravitational effects are negligible on the current carrying ring.
- The conservation of magnetic potential ( U ) energy in the form of Kinetic energy ( Ek ) is valid for the given application:
Ui + Eki = Uf + Ekf
Where,
Eki : The initial kinetic energy ( initially at rest ) = 0
Ekf : The final kinetic energy at second position
- The loss in potential energy stored is due to the conversion of potential energy into rotational kinetic energy of current carrying ring.
-ΔU = Ekf
0.5*T*w^2 = -ΔU
w^2 = -ΔU*2 / T
Where,
w: The angular speed at second position
w = √(0.000747871*2 / 6.50×10^−7)
w = 47.97 rad / s
Taking into account the rule of three for the change of units, the mass of the book is 45600 miligrams.
First of all, the rule of three is a mathematical tool that helps you quickly solve proportionality problems.
Having three known values and one unknown, a proportional relationship is established between all of them in order to find the fourth term of the proportion.
If the relationship between the magnitudes is direct (when one magnitude increases, so does the other; or when one magnitude decreases, so does the other), the rule of three is applied as follows, where a, b and c are known values and x is the unknown to calculate:
a → b
c → x
So:
Being 1 kg equivalent to 1000000 milligrams, In this case the rule of three is applied as follows: if 1 kg equals 1000000 milligrams, 4.56×10⁻² kg equals how many milligrams?
1 kg → 1000000 milligrams
4.56×10⁻² kg → x
So:
<u><em>x=45600 miligrams</em></u>
In summary, the mass of the book is 45600 miligrams.
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