TLDR: It will reach a maximum when the angle between the area vector and the magnetic field vector are perpendicular to one another.
This is an example that requires you to investigate the properties that occur in electric generators; for example, hydroelectric dams produce electricity by forcing a coil to rotate in the presence of a magnetic field, generating a current.
To solve this, we need to understand the principles of electromotive forces and Lenz’ Law; changing the magnetic field conditions around anything with this potential causes an induced current in the wire that resists this change. This principle is known as Lenz’ Law, and can be described using equations that are specific to certain situations. For this, we need the two that are useful here:
e = -N•dI/dt; dI = ABcos(theta)
where “e” describes the electromotive force, “N” describes the number of loops in the coil, “dI” describes the change in magnetic flux, “dt” describes the change in time, “A” describes the area vector of the coil (this points perpendicular to the loops, intersecting it in open space), “B” describes the magnetic field vector, and theta describes the angle between the area and mag vectors.
Because the number of loops remains constant and the speed of the coils rotation isn’t up for us to decide, the only thing that can increase or decrease the emf is the change in magnetic flux, represented by ABcos(theta). The magnetic field and the size of the loop are also constant, so all we can control is the angle between the two. To generate the largest emf, we need cos(theta) to be as large as possible. To do this, we can search a graph of cos(theta) for the highest point. This occurs when theta equals 90 degrees, or a right angle. Therefore, the electromotive potential will reach a maximum when the angle between the area vector and the magnetic field vector are perpendicular to one another.
Hope this helps!
Answer:
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Explanation:
Answer:
W=315 x 10⁵ J
Explanation:
Given that
F= 2.5 x 10⁵ N
d= 90 m
K.E.=5.4 x 10⁷ J
We know that work done by all force is equal to the change in kinetic energy
Lets take work done by catapult is W
W + F.d= K.E.
W= 5.4 x 10⁷ - 2.5 x 10⁵ x 90 J
W= (540 - 25 x 9) 10⁵ J
W=315 x 10⁵ J
Answer:
a) 
For this case we know the following values:




So then if we replace we got:

b) 
With 
And replacing we have:

And then the scattered wavelength is given by:

And the energy of the scattered photon is given by:

c) 
Explanation
Part a
For this case we can use the Compton shift equation given by:
For this case we know the following values:
So then if we replace we got:
Part b
For this cas we can calculate the wavelength of the phton with this formula:
With
And replacing we have:
And then the scattered wavelength is given by:
And the energy of the scattered photon is given by:
Part c
For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:
Answer:
Of course harder
Explanation:
Just imagone the floor is wet and you walk on it, do u feel it hard or easy to walk? :D