Answer:
λ = 451.7 nm
Explanation:
The expression for the constructive interference of the double diffraction experiment is
d sin θ = m λ
let's use trigonometry
tan θ = y / L
how the experiment occurs at very small angles
tan θ = sin θ / cos θ = sin θ
sin θ = y / L
we substitute
d y / L = m λ
λ =
let's calculate
λ =
λ = 4.51699 10⁻⁷ m
λ = 4.517 10⁻⁷ m (109 nm / 1m)
λ = 451.7 nm
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
Answer:
Explanation:
Given
Minute hand length =16 cm
Time at a quarter after the hour to half past i.e. 1 hr 45 min
Angle covered by minute hand in 1 hr is 360 and in 45 minutes 270
(c)For the next half hour
Effectively it has covered 2 revolution and a quarter
angle turned
(f)Hour after that
After an hour it again comes back to its original position thus displacement is same =25.136
Angle turned will also be same i.e.
Answer:
a) In the new transformer there are 42 turns in the secondary per turn in the primary, while in the old transformer there were 32 turns per turn in the primary.
b) The new output is 86% of the old output
c) The losses in the new line are 74% the losses in the old line.
Explanation:
a) To relate the turns of primary and secondary to the ratio of voltage we have this expression:
In the old transformer the ratio of voltages was:
In the new transformer the ratio of voltages is:
In the new transformer there are 42 turns in the secondary per turn in the primary, while in the old transformer there were 32 turns per turn in the primary.
b) The new current ratio is
If the old current output was 425 kV, the ratio of current was:
Then, the ratio of the new output over the old output is:
The new output is 86% of the old output (smaller output currents lower the losses on the transmission line).
c) The power loss is expressed as:
Then, the ratio of losses is (R is constant for both power losses):
The losses in the new line are 74% the losses in the old line.