A circular loop of wire has radius 7.80cm . A sinusoidal electromagnetic plane wave traveling in air passes through the loop, with the direction of the magnetic field of the wave perpendicular to the plane of the loop. The intensity of the wave at the location of the loop is 2.03�10?2W/m2 , and the wavelength of the wave is 6.20m .What is the maximum emf induced in the loop? Express your answer with the appropriate units.I stumbled through the formulas I do know for EMF but cant seem to figure out how to get the right answer. Please help and provide explanation! Thanks
Answers
Answer:
fem = - 4.50 10²² V
Explanation:
For the solution of this problem we must use the equation of the induced electromotive force or Faraday's law
E = - d Φ / dt = d (BA cos θ) dt
In this case they tell us that the magnetic field is perpendicular to the plane of the loop, as the normal to the surface of the loop is in the direction of the radius, the angle between the field and this normal is zero, so cos 0º = 1. The area of the loop is constant, with this the equation is
E = - A dB / dt (1)
To find field B, we have the relationships of electromagnetic waves
E = c B
The intensity or poynting vector for the wave is described by the equation
S = I = 1 / μ₀ E x B = 1 /μ₀ E B
We replace
I = 1 /μ₀ (cB) B = c /μ₀ B²
This is the instantaneous intensity.
B = √ (μ₀ I /c)
We substitute in equation 1
E = - A μ₀/c d I / dt
With the maximum value we are asked to change it derived from variations
E = -A c/μ₀ ΔI / Δt
It remains to find the time of the variation. Let's use the equation
c = λ f = λ / T
T = λ / c
T = 6.20 / 3 10⁸
T = 2.06 10⁻⁸ s
We already have all the values to calculate the fem
fem = - π r² c/μ₀ ΔI/Δt
fem = - (π 0.078²) (3 10⁸/(4π 10⁻⁷) (2.03 10² -0) / (2.06 10⁻⁸ - 0)
fem = - 4.50 10²² V
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Answers
The force between objects that are any distance apart is expressed as
According to the gravitational law, the force acting on an object is directly proportional to the product of their masses and inversely proportional to the square of their distance apart. Mathematically,
M and m are the masses
r is the distance between the masses
If the force between objects that are 10 meters apart, hence;
To find the force between objects that are any distance apart, we will use the same formula above to have;
Substitute the result above into the expression to have:
Hence the force between objects that are any distance apart is expressed as
Learn more on gravitational law here: brainly.com/question/11760568
Answer:
F' = 100 F/r²
Explanation:
The gravitational force of attraction between two objects is given by the Newton's Gravitational Formula. The Newton's Gravitational Formula is as follows:
F = Gm₁m₂/r²
where,
F = Force between objects
G = Universal Gravitational Constant
m₁ = mass of first object
m₂ = mass of second object
r = distance between objects = 10 m
Therefore,
F = Gm₁m₂/10²
Gm₁m₂ = 100F --------------------- equation (1)
Now, we consider these objects at any distance r apart. So, the force becomes:
F' = Gm₁m₂/r²
using equation (1), we get:
F' = 100 F/r²
So, if the force (F) between objects 10 m apart is known, we can find it at any distance from the above formula.
Answers
Answer:
Please find the attached file for the figure.
Explanation:
Given that a bicyclist speeds along a road at 10 m/s for 6 seconds.
Its acceleration = 10/6 = 1.667 m/s^2
The distance covered = 1/2 × 10 × 6
Distance covered = 30 m
That is, displacement = 30 m
Then she stops for three seconds to make a 180˚ turn and then travels at 5 m/s for 3 seconds.
The acceleration = 5/3 = 1.667 m/s^2
The displacement = 1/2 × 5 × 3
Displacement = 7.5 m
The resultant acceleration will be equal to zero.
While the resultant displacement will be:
Displacement = 30 - 7.5 = 22.5 m
Please find the attached file for the sketch.
Answers
Answer:
1kg
Explanation:
this box is the smallest and weighs the least. Hope this helps :]
Answers
Answer: C = 1.319×10^-11 F
Explanation: The formulae that relates the capacitance of a capacitor to the area of the plates, distance between the plates and dielectric constant is given as
C = kε0A/d
Where C = capacitance of plates =?
k = dielectric constant = 2.0
Area of plates = 16.4cm² = 0.00164 m²
d = distance between plates = 2.2 mm = 0.0022m
By substituting the parameters, we have that
C = 2 × 8.85×10 ^-12 ×0.00164/ 0.0022
C = 0.029028 × 10^-12/ 0.0022
C = 13.19× 10^-12
C = 1.319×10^-11 F
Answers
Answer:
U/U₀ = 2
(factor of 2 i.e U = 2U₀)
Therefore, the energy stored in the capacitor is doubled when the plate separation is doubled while the capacitor has been disconnected
Explanation:
Energy stored in a capacitor can be expressed as;
U = 0.5CV^2 = Q^2/2C
And
C = ε₀ A/d
Where
C = capacitance
V = potential difference
Q = charge
A = Area of plates
d = distance between plates
So
U = Q^2/2C = dQ^2/2ε₀ A
The initial energy of the capacitor at d = d₀ is
U₀ = Q^2/2C = d₀Q^2/2ε₀ A ....1
When the plate separation is increased after the capacitor has been disconnected, the charge Q of the capacitor remain constant.
The final energy stored in the capacitor at d = 2d₀ is
U = 2d₀Q^2/2ε₀ A ...2
The factor U/U₀ can be derived by substituting equation 1 and 2
U/U₀ = (2d₀Q^2/2ε₀ A)/( d₀Q^2/2ε₀ A )
Simplifying we have;
U/U₀ = 2
U = 2U₀
Therefore, the energy stored in the capacitor is doubled when the plate separation is doubled while the capacitor has been disconnected.
Answers
Answer:
The recoil velocity of the raft is 1.205 m/s.
Explanation:
given that,
Mass of the swimmer,
Mass of the raft,
Velocity of the swimmer, v = +4.6 m/s
It is mentioned that the swimmer then runs off the raft, the total linear momentum of the swimmer/raft system is conserved. Let V is the recoil velocity of the raft.
V = -1.205 m/s
So, the recoil velocity of the raft is 1.205 m/s. Hence, this is the required solution.
Answer:
The recoil velocity of the raft would be (pointing to the left if the swimmer runs to the right)
Explanation:
The problem states thatthe swimmer has a mass of m=55 kg, and the raft has a mass of M=210 kg. Then, it says that the swimmer runs off the raft with a (final) velocity of v=4.6 m/s relative to the shore.
To analyze it, we take a system of "two particles", wich means that we will consider the swimmer and the raft as a hole system, aisolated from the rest of the world.
Then, from the shore, we can put our reference system and take the initial moment when the swimmer and the raft are stationary. This means that the initial momentum is equal to zero:
Besides, we can use magnitudes instead of vectors because the problem will develope in only one dimension after the initial stationary moment (x direction, positive to the side of the running swimmer, and negative to the side of the recoling raft), this means that we can write the final momentum as
The final momentum is equal to zero due to conservation of momentum (because there are no external forces in the problem, for the system "swimmer-raft"), so the momentum is constant.
Then, from that previous relation we can clear
wich is the recoil velocity of the raft, and it is pointing to the left (we established this when we said that the raft was going to the negative side of the system of reference, and when we put a minus in the raft term inside the momentum equation).