Answer: Astro 1 will have a 10x larger field of view than the Hubble.
Explanation: The hubble will also be extremely light weight that way it can go further into space and the mission will be able to last a longer amount of time.
The difference between a DC and an AC generator is that
a. the DC generator has one unbroken slip ring.
b. the AC generator has one unbroken slip ring
c. the DC generator has one slip ring splitin two halves.
d. the AC generator has one slip ring split in two halves.
e The DC generator has twounbroken sip rings
Answer:
The AC generator has one unbroken slip ring
Explanation:
In physics, the application of electromagnetic induction can be seen in generators and dynamos. Electromagnetic induction is the process of generating electricity using magnets. It found applications in generators and the types of generator they found application is in AC and DC generator.
An AC generator is also called a Dynamo. A DC generator contains what is called a SPLIT RING fixed to the end of the coil which can be separated and coupled back according to the name "split". An AC generator also called a Dynamo makes use of a SLIP ring which cannot be divided into two. It comes as an entity. The presence of this rings is what differentiates a DC generator from an AC generator.
We can replace split rings with slip rings when converting a DC generator to an AC generator and vice versa.
It can therefore be concluded that the difference between a DC and an AC generator is that the AC generator has one unbroken slip ring.
A 3200-lb car is moving at 64 ft/s down a 30-degree grade when it runs out of fuel. Find its velocity after that if friction exerts a resistive force with magnitude proportional to the square of the speed with k
Answer:
The velocity is 40 ft/sec.
Explanation:
Given that,
Force = 3200 lb
Angle = 30°
Speed = 64 ft/s
The resistive force with magnitude proportional to the square of the speed,
[tex]F_{r}=kv^2[/tex]
Where, k = 1 lb s²/ft²
We need to calculate the velocity
Using balance equation
[tex]F\sin\theta-F_{r}=m\dfrac{d^2v}{dt^2}[/tex]
Put the value into the formula
[tex]3200\sin 30-kv^2=m\dfrac{d^2v}{dt^2}[/tex]
Put the value of k
[tex]3200\times\dfrac{1}{2}-v^2=m\dfrac{d^2v}{dt^2}[/tex]
[tex]1600-v^2=m\dfrac{d^2v}{dt^2}[/tex]
At terminal velocity [tex]\dfrac{d^2v}{dt^2}=0[/tex]
So, [tex]1600-v^2=0[/tex]
[tex]v=\sqrt{1600}[/tex]
[tex]v=40\ ft/sec[/tex]
Hence, The velocity is 40 ft/sec.
If you wanted to make your own lenses for a telescope, what features of a lens do you think would affect the images that you can see
Answer:
Therefore the characteristics to be found are:
* the focal length must be large and the focal length of the eyepiece must be small
* The diameter of the objective lens should be as large as possible, to be able to collect small without need from light
* The system must be configured to the far sight tip,
Explanation:
The length of the telescope is
L = f_ocular + f_objetive
the magnification of the telescope is
m = - f_objective / f_ocular
These are the two equations that describe the behavior of the telescope. Therefore the characteristics to be found are:
* the focal length must be large and the focal length of the eyepiece must be small
* The diameter of the objective lens should be as large as possible, to be able to collect small without need from light
* The system must be configured to the far sight tip,
Electrons are accelerated through a voltage difference of 270 kV inside a high voltage accelerator tube. What is the final kinetic energy of the electrons?
Each electron winds up with kinetic energy of
(270 keV)
plus
(whatever KE it had when it started accelerating).
A semi-circular loop consisting of one turn of wire is place in the x-y plane. A constant magnetic field B=1.7T points along the negative z-axis(into the page), and a current I=0.7A flows counterclockwisefrom the positive z-axis. The net magnetic force on the circular section of the loop points in what direction? What is the net magnetice force on the circular section of the loop?
Answer:
The direction of net magnetic force on the circular section of the loop is in the positive y-axis
The net magnetic force on the circular section of the loop is 3.74 N
Explanation:
The magnetic field strength [tex]B[/tex] = 1.7 T
the current [tex]I[/tex] = 0.7 A
The diameter of the loop = 2 m
the length of the circular section of the semi-circular loop [tex]l[/tex] = πd/2
==> [tex]l[/tex] = (3.142 x 2)/2 = 3.142 m
The force on the semi-circular is given as
F = [tex]BIl[/tex] sin ∅
but the loop is perpendicular to the field, therefore
sin ∅ = sin 90° = 1
F = 1.7 x 0.7 x 3.142 x 1 = 3.74 N
The right hand rule states that "if the fingers of the right hand are held parallel to each other in the direction of the magnetic field, and the thumb is held at right angle to the other fingers in the direction of the flow of current. The palm will push in the direction of the magnetic force on the conductor".
According to the right hand rule, the direction of net magnetic force on the circular section of the loop is in the positive y-axis
¿Qué resistencia debe ser conectada en paralelo con una de 20 Ω para hacer una
resistencia combinada de 15 Ω?
Answer:
60 Ω
Explanation:
R(com) = 15 Ω
1/R(com) = 1/R1 + 1/R2 + 1/R3 ..... + 1/Rn
1/15 = 1/20 + 1/R2
1/R2 = 1/15 - 1/20
1/R2 = (4 - 3) / 60
1/R2 = 1/60
R2 = 60 Ω
así, la combinada de resistencia necesaria es 60 Ω
Two wires carry current I1 = 73 A and I2 = 31 A in the opposite directions parallel to the x-axis at y1 = 3 cm and y2 = 13 cm. Where on the y-axis (in cm) is the magnetic field zero?
Answer:
The position on the y-axis where the magnetic field is zero is at y = 10 cm
Explanation:
The magnetic field B due to a long straight wire carrying a current, i at a distance R from the wire is given by
B = μ₀i/2πR
Now, let y be the point where the magnetic fields of both wires are equal.
So, the magnetic field due to wire 1 carrying current i₁ = 73 A is
B₁ = μ₀i₁/2π(y - 3) and
the magnetic field due to wire 2 carrying current i₂ = 31 A is
B₂ = μ₀i₂/2π(13 - y)
At the point where the magnetic field is zero, B₁ = B₂. So,
μ₀i₁/2π(y - 3) = μ₀i₂/2π(13 - y)
cancelling out μ₀ and 2π, we have
i₁/(x - y) = i₂/(13 - y)
cross-multiplying, we have
(13 - y)i₁ = (y - 3)i₂
Substituting the values of i₁ and i₂, we have
(13 - y)73 = (y - 3)31
949 - 73y = 31y - 93
Collecting like terms, we have
949 + 93 = 73y + 31y
1042 = 104y
dividing through by 104, we have
y = 1042/104
y = 10.02 cm
y ≅ 10 cm
So, the position on the y-axis where the magnetic field is zero is at y = 10 cm
Light of wavelength 520 nm is used to illuminate normally two glass plates 21.1 cm in length that touch at one end and are separated at the other by a wire of radius 0.028 mm. How many bright fringes appear along the total length of the plates.
Answer:
The number is [tex]Z = 216 \ fringes[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 520 \ nm = 520 *10^{-9} \ m[/tex]
The length of the glass plates is [tex]y = 21.1cm = 0.211 \ m[/tex]
The distance between the plates (radius of wire ) = [tex]d = 0.028 mm = 2.8 *10^{-5} \ m[/tex]
Generally the condition for constructive interference in a film is mathematically represented as
[tex]2 * t = [m + \frac{1}{2} ]\lambda[/tex]
Where t is the thickness of the separation between the glass i.e
t = 0 at the edge where the glasses are touching each other and
t = 2d at the edge where the glasses are separated by the wire
m is the order of the fringe it starts from 0, 1 , 2 ...
So
[tex]2 * 2 * d = [m + \frac{1}{2} ] 520 *10^{-9}[/tex]
=> [tex]2 * 2 * (2.8 *10^{-5}) = [m + \frac{1}{2} ] 520 *10^{-9}[/tex]
=>
[tex]m = 215[/tex]
given that we start counting m from zero
it means that the number of bright fringes that would appear is
[tex]Z = m + 1[/tex]
=> [tex]Z = 215 +1[/tex]
=> [tex]Z = 216 \ fringes[/tex]
You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. You set your slit spacing at 1.11 mm and place your screen 8.63 m from the slits. Then, you illuminate the slits with your new toy and find on the screen that the tenth bright fringe is 4.71 cm away from the central bright fringe (counted as the zeroth bright fringe). What is your laser's wavelength lambda expressed in nanometers?
Answer:
λ = 605.80 nm
Explanation:
These double-slit experiments the equation for constructive interference is
d sin θ = m λ
where d is the distance between the slits, λ the wavelength of light and m an integer that determines the order of interference.
In this case, the distance between the slits is d = 1.11 mm = 1.11 10⁻³ m, the distance to the screen is L = 8.63 m, the range number m = 10 and ay = 4.71 cm
Let's use trigonometry to find the angle
tan θ = y / L
as the angles are very small
tan θ = sin θ / cos θ = sin θ
we substitute
sin θ = y / L
we substitute in the first equation
d y / L = m λ
λ = d y / m L
let's calculate
λ = 1.11 10⁻³ 4.71 10⁻²/ (10 8.63)
λ = 6.05805 10⁻⁷ m
let's reduce to nm
λ = 6.05805 10⁻⁷ m (10⁹ nm / 1m)
λ = 605.80 nm
Suppose you observed the equation for a traveling wave to be y(x, t) = A cos(kx − ????t), where its amplitude of oscillations was 0.15 m, its wavelength was two meters, and the period was 2/15 s. If a point on the wave at a specific time has a displacement of 0.12 m, what is the transverse speed of that point?
Answer:
15m/sExplanation:
The equation for a traveling wave as expressed as y(x, t) = A cos(kx − [tex]\omega[/tex]t) where An is the amplitude f oscillation, [tex]\omega[/tex] is the angular velocity and x is the horizontal displacement and y is the vertical displacement.
From the formula; [tex]k =\frac{2\pi x}{\lambda} \ and \ \omega = 2 \pi f[/tex] where;
[tex]\lambda \ is\ the \ wavelength \ and\ f \ is\ the\ frequency[/tex]
Before we can get the transverse speed, we need to get the frequency and the wavelength.
frequency = 1/period
Given period = 2/15 s
Frequency = [tex]\frac{1}{(2/15)}[/tex]
frequency = 1 * 15/2
frequency f = 15/2 Hertz
Given wavelength [tex]\lambda[/tex] = 2m
Transverse speed [tex]v = f \lambda[/tex]
[tex]v = 15/2 * 2\\\\v = 30/2\\\\v = 15m/s[/tex]
Hence, the transverse speed at that point is 15m/s
A girl is sitting on the edge of a pier with her legs dangling over the water. Her soles are 80.0 cm above the surface of the water. A boy in the water looks up at her feet and wants to touch them with a reed. (nwater =1.333). He will see her soles as being:____
a. right at the water surface.
b. 53.3 cm above the water surface.
c. exactly 80.0 cm above the water surface.
d. 107 cm above the water surface.
e. an infinite distance above the water surface.
Answer:
d. 107 cm above the water surface.
Explanation:
The refractive index of water and air = 1.333
The real height of the girl's sole above water = 80.0 cm
From the water, the apparent height of the girl's sole will be higher than it really is in reality by a factor that is the refractive index.
The boy in the water will therefore see her feet as being
80.0 cm x 1.333 = 106.64 cm above the water
That is approximately 107 cm above the water
As the frequency of the ac voltage across a capacitor approaches zero, the capacitive reactance of that capacitor:_______.
a. approaches zero.
b. approaches infinity.
c. approaches unity.
d. none of the above.
Answer:
b. approaches infinity
Explanation:
Because Capacitive reactance is given as Xc = 1/ωC
So we can see that the value of capacitive reactance and therefore its overall impedance (in Ohms) decreases to zero as the frequency increases acting like a short circuit.
Same as the frequency approaches zero or DC, the capacitors reactance increases to infinity, acting like an open circuit which is why capacitors block DC
Question 18(Multiple Choice Worth 2 polnis)
When riding your skateboard you crash into a curb, the skateboard stops, and you continue moving forward. Which law of
motion is being described in this scenario?
O Law of Universal Gravitation
o Newton's Second Law of Motion
o Law of Conservation of Energy
o Newton's First Law of Motion
3. Which of the following accurately describes circuits?
O A. In a parallel circuit, the same amount of current flows through each part of the circuit
O B. In a series circuit, the amount of current passing through each part of the circuit may vary
O C. In a series circuit, the current can flow through only one path from start to finish
O D. In a parallel circuit, there's only one path for the current to travel.
Answer:
Option (c)
Explanation:
In a Series circuit, as the components are connected end-to-end ,the current can flow through only one path from start to finish.
(C.) is the only correct statement in the list of choices.
In a series circuit, the current can flow through only one path from start to finish.
The advantage of a hydraulic lever is A : it transforms a small force acting over a large distance into a large force acting over a small distance. B : it transforms a small force acting over a small distance into a large force acting over a large distance. C : it allows you to exert a larger force with less work. D : it transforms a large force acting over a large distance into a small force acting over a small distance. E : it transforms a large force acting over a small distance into a small force acting over a large distance.
Answer:
A) it transforms a small force acting over a large distance into a large force acting over a small distance.
Explanation:
The hydraulic lever works based on Pascal's law of transmission of pressure through a fluid. In the hydraulic lever, the pressure transmitted is the same.
Pressure transmitted P = F/A
where F is the force applied
and A is the area over which the force is applied.
This pressure can be manipulated on the input end as a small force applied over a small area, and then be transmitted to the output end as a large force over a large area.
F/A = f/a
where the left side of the equation is for the output, and the right side is for the input.
The volume of the displaced fluid will be the same on both ends of the hydraulic lever. Since we know that
volume V = (area A) x (distance d)
this means that the the piston on the input smaller area of the hydraulic lever will travel a greater distance, while the piston on the larger output area of the lever will travel a small distance.
From all these, we can see that the advantage of a hydraulic lever is that it transforms a small force acting over a large distance into a large force acting over a small distance.
The speed of sound through air is 340 m/s. If a person hears the clap of thunder 9.6 s after seeing the bolt of lightning, how far away is the lightning?
Explanation:
Distance = speed × time
d = (340 m/s) (9.6 s)
d = 3264 m
A small helium-neon laser emits red visible light with a power of 5.40 mW in a beam of diameter 2.30 mm.
Required:
a. What is the amplitude of the electric field of the light? Express your answer with the appropriate units.
b. What is the amplitude of the magnetic field of the light?
c. What is the average energy density associated with the electric field? Express your answer with the appropriate units.
d. What is the average energy density associated with the magnetic field? Express your answer with the appropriate units.
E) What is the total energy contained in a 1.00-m length of the beam? Express your answer with the appropriate units.
Answer:
A. 990v/m
B.330x10^-8T
C.2.19x10^-6J/m³
D.1.45x10^-11J
Explanation:
See attached file
Light passes from a material with index of refraction 1.3 into one with index of refraction 1.2. Compared to the incident ray, what happens to the refracted ray?
Answer:
It bends away from the normal
Explanation:
From Snell's law of Refraction, when a ray passes from a medium of lower Refractive index to a medium with higher Refractive index, the Refractive ray will bend towards the normal. However, when the ray passes from a medium of higher Refractive index to a medium of lower Refractive index, the Refractive ray will bend away from the normal.
Now, from the question we are told that Light passes from a material with index of refraction 1.3 into one with index of refraction 1.2.
This means from a higher Refractive index to a lower one and from Snell's law as earlier said, the refracted ray will bend away from the normal
The refracted ray is seen to bend away from the normal.
Let us recall that an optically denser medium will have a higher refractive index. This means that the medium with a refractive index of 1.3 is the denser medium and the medium with a refractive index of 1.2 is the less dense medium.
From the statement in the question, we can boldly say that light is travelling from a denser to less dense medium given the values of the refractive index given. When light is travelling from a denser to a less dense medium, the refracted ray bends away from the normal.
Learn more: https://brainly.com/question/1445383
"A thin film with an index of refraction of 1.50 is placed in one of the beams of a Michelson interferometer. If this causes a shift of 8 bright fringes in the pattern produced by light of wavelength 540 nm, what is the thickness of the film?"
Answer:
The film thickness is 4.32 * 10^-6 m
Explanation:
Here in this question, we are interested in calculating the thickness of the film.
Mathematically;
The number of fringes shifted when we insert a film of refractive index n and thickness L in the Michelson Interferometer is given as;
ΔN = (2L/λ) (n-1)
where λ is the wavelength of the light used
Let’s make L the subject of the formula
(λ * ΔN)/2(n-1) = L
From the question ΔN = 8 , λ = 540 nm, n = 1.5
Plugging these values, we have
L = ((540 * 10^-9 * 8)/2(1.5-1) = (4320 * 10^-9)/1 = 4.32 * 10^-6 m
With the same block-spring system from above, imagine doubling the displacement of the block to start the motion. By what factor would the following change?
A. Kinetic energy when passing through the equilibrium position.
B. Speed when passing through the equilibrium position.
Answer:
A) K / K₀ = 4 b) v / v₀ = 4
Explanation:
A) For this exercise we can use the conservation of mechanical energy
in the problem it indicates that the displacement was doubled (x = 2xo)
starting point. At the position of maximum displacement
Em₀ = Ke = ½ k (2x₀)²
final point. In the equilibrium position
[tex]Em_{f}[/tex] = K = ½ m v²
Em₀ = Em_{f}
½ k 4 x₀² = K
(½ K x₀²) = K₀
K = 4 K₀
K / K₀ = 4
B) the speed value
½ k 4 x₀² = ½ m v²
v = 4 (k / m) x₀
if we call
v₀ = k / m x₀
v = 4 v₀
v / v₀ = 4
An electrostatic paint sprayer contains a metal sphere at an electric potential of 25.0 kV with respect to an electrically grounded object. Positively charged paint droplets are repelled away from the paint sprayer's positively charged sphere and towards the grounded object. What charge must a 0.168-mg drop of paint have so that it will arrive at the object with a speed of 18.8 m/s
Answer:
The charge is [tex]Q = 2.177 *10^{-9} \ C[/tex]
Explanation:
From the question we are told that
The electric potential is [tex]V = 25.0 \ kV = 25.0 *10^{3}\ V[/tex]
The mass of the drop is [tex]m = 0.168 \ m g = 0.168 *10^{-3} \ g = 0.168 *10^{-6}\ kg[/tex]
The speed is [tex]v = 18.8 \ m/s[/tex]
Generally the charge on the paint drop due to the electric potential which will give it the speed stated in the question is mathematically represented as
[tex]Q = \frac{m v^2 }{ 2 * V }[/tex]
Substituting values
[tex]Q = \frac{0.168 *10^{-6} (18)^2 }{ 2 * 25*10^3 }[/tex]
[tex]Q = 2.177 *10^{-9} \ C[/tex]
7. A sound wave begins traveling through a thin metal rod at one end with a speed that is 15 times the speed of sound in air. If an observer at the other end of the rod hears the sound twice, one from the sound traveling through the rod and one from the sound traveling through the air, with a time delay of 0.12 s, how long is the rod? The speed of sound in air is 343 m/s.
Answer:
L = 44,096 m
Explanation:
The speed of the sound wave is constant therefore we can use the relations of uniform kinematics
v = x / t
the speed of the wave in the bar is
v = 15 v or
v = 15 343
v = 5145 m / s
The sound at the bar goes the distance
L = v t
Sound in the air travels the same distance
L = v_air (t + 0.12)
as the two recognize the same dissonance,
v t = v_air (t +0.12)
t (v- v_air) = 0.12 v_air
t = 0.12 v_air / (v -v_air)
l
et's calculate
t = 0.12 343 / (5145 - 343)
t = 8.57 10-3 s
The length of the bar is
L = 5145 8.57 10-3
L = 44,096 m
A city of Punjab has a 15 percent chance of wet weather on any given day. What is the probability that it will take a week for it three wet weather on 3 separate days? Also find its Standard Deviation
Answer:
so the probability will be = 0.062
Standard deviation = 0.8925
Explanation:
The probability of rain = 15% = 15/100= 0.15
and the probability of no rain=q = 1-p= 1-0.15= 0.85
The number of trials = 7
so the probability will be
7C3 * ( 0.15)^3 (0.85)^4= 35* 0.003375 * 0.52200 =0.06166= 0.062
Taking this as binomial as the p and q are constant and also the trials are independent .
For a binomial distribution
Standard deviation = npq= 0.15 *0.85 *7= 0.8925
A car moving at 36 m/s passes a stationary police car whose siren has a frequency of 500 Hz. What is the change in the frequency (in Hz) heard by an observer in the moving car as he passes the police car? (The speed of sound in air is 343 m/s.)
Answer:
Change in the frequency (in Hz) = 104.96 Hz
Explanation:
Given:
Speed of sound in air (v) = 343 m/s
Speed of car (v1) 36 m/s
Frequency(f) = 500 Hz
Find:
Change in the frequency (in Hz)
Computation:
Frequency hear by the observer(before)(f1) = [f(v+v1)] / v
Frequency hear by the observer(f1) = [500(343+36)] / 343
Frequency hear by the observer(f1) = 552.48 Hz
Frequency hear by the observer(after)(f2) = [f(v-v1)] / v
Frequency hear by the observer(f2) = [500(343-36)] / 343
Frequency hear by the observer(f2) = 447.52 Hz
Change in the frequency (in Hz) = f1 - f2
Change in the frequency (in Hz) = 552.48 Hz - 447.52 Hz
Change in the frequency (in Hz) = 104.96 Hz
If you have a density of 100 kg/L, and a mass of 1000 units, tell me the following: First what are the mass units?
Answer:
The mas unit is the the 'Kilogram' written as 'kg'
Volume is 10 L
Explanation:
The complete question is
If you have a density of 100 kg/L, and a mass of 1000 units, tell me the following: First, what are the mass units?
Second, what is the volume
mass units is the 'Kilogram', written as 'kg'
density = mass/volume = 100 kg/L
the mass = 1000 kg
volume = mass/density = 1000/100 = 10 L
The accommodation limits for a nearsighted person's eyes are 20.0 cm and 82.0 cm. When he wears his glasses, he can see faraway objects clearly. At what minimum distance is he able to see objects clearly
Answer;
26.45cm
See attached file for explanation
Equipotential lines are lines with equal electric potential (for example, all the points with an electric potential of 5.0 V). Using the plot tool that comes with voltmeter (pencil icon) make two equipotential lines at r = 0.5 m and r = 1.5 m. Enable electric field vectors in the simulation. Put an electric field sensor at different points on the equipotential line and note the direction of the electric field vector. What can you conclude about the direction of the electric field vector in relation to the equipotential lines?
The direction for each field vector is perpendicular to equipotential lines.
Take a snapshot of the simulation showing equipotential lines and paste to a word document.
....................
ametal of mass 0.6kg is heated by an electric heater connected to 15v batter when the ammeter reading is 3A its tempeeature rises feom 20c to 85c in 10 minutes calculate the s.h.c of metal cylinder
Answer:
692 J/kg/°C
Explanation:
Electric energy added = amount of heat
Power × time = mass × SHC × increase in temperature
Pt = mCΔT
(15 V × 3 A) (10 min × 60 s/min) = (0.6 kg) C (85°C − 20°C)
C = 692 J/kg/°C
Two football teams, the Raiders and the 49ers are engaged in a tug-of-war. The Raiders are pulling with a force of 5000N. Which of the following is an accurate statement?
A. The tension in the rope depends on whether or not the teams are in equilibrium.
B. The 49ers are pulling with a force of more than 5000N because of course they’d be winning.
C. The 49ers are pulling with a force of 5000N.
D. The tension in the rope is 10,000N.
E. None of these statements are true.
Answer:
E. None of these statements are true.
Explanation:
We can't say the exact or approximate amount of tension on the rope, since we do know for sure from the statement who is winning.
for A, the tension on the rope does not depend on if both teams pull are in equilibrium.
for B, the 49ers would be pulling with a force more than 5000 N, if they were winning. The problem is that we can't say with all confidence that they'd be winning.
for C, we don't know how much tension exists on the rope, and its direction, so we can't work out how much tension the 49ers are pulling the rope with.
for D, just as for C above, we can't work out how much tension there is on the rope, since we do not know how much force the 49ers are pulling with.
we go with option E.
Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J = cr2 = 9.00 ✕ 106 A/m4 r2. What is the current (in A) through the inner section of the wire from the center to r = 0.5R?
Answer:
The current is [tex]I = 8.9 *10^{-5} \ A[/tex]
Explanation:
From the question we are told that
The radius is [tex]r = 3.17 \ mm = 3.17 *10^{-3} \ m[/tex]
The current density is [tex]J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2[/tex]
The distance we are considering is [tex]r = 0.5 R = 0.001585[/tex]
Generally current density is mathematically represented as
[tex]J = \frac{I}{A }[/tex]
Where A is the cross-sectional area represented as
[tex]A = \pi r^2[/tex]
=> [tex]J = \frac{I}{\pi r^2 }[/tex]
=> [tex]I = J * (\pi r^2 )[/tex]
Now the change in current per unit length is mathematically evaluated as
[tex]dI = 2 J * \pi r dr[/tex]
Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows
[tex]I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr[/tex]
[tex]I = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr[/tex]
[tex]I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.[/tex]
[tex]I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ][/tex]
substituting values
[tex]I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ][/tex]
[tex]I = 8.9 *10^{-5} \ A[/tex]
