A uniform disk a uniform hoop and a uniform sphere are released at the same time at the top of an inclined ramp. They all roll w
1 answer:
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(a) The net flux through the coil is zero.
In fact, the magnetic field generated by the wire forms concentric circles around the wire. The wire is placed along the diameter of the coil, so we can imagine as it divides the coil into two emisphere. Therefore, the magnetic field of the wire is perpendicular to the plane of the coil, but the direction of the field is opposite in the two emispheres. Since the two emispheres have same area, then the magnetic fluxes in the two emispheres are equal but opposite in sign, and so they cancel out when summing them together to find the net flux.
(b) If the wire passes through the center of the coil but it is perpendicular to the plane of the wire, the net flux through the coil is still zero.
In fact, the magnetic field generated by the wire forms concentric lines around the wire, so it is parallel to the plane of the coil. But the flux is equal to
where
is the angle between the direction of the magnetic field and the perpendicular to the plane of the coil, so in this case 
and so the cosine is zero, therefore the net flux is zero.
In fact, the magnetic field generated by the wire forms concentric circles around the wire. The wire is placed along the diameter of the coil, so we can imagine as it divides the coil into two emisphere. Therefore, the magnetic field of the wire is perpendicular to the plane of the coil, but the direction of the field is opposite in the two emispheres. Since the two emispheres have same area, then the magnetic fluxes in the two emispheres are equal but opposite in sign, and so they cancel out when summing them together to find the net flux.
(b) If the wire passes through the center of the coil but it is perpendicular to the plane of the wire, the net flux through the coil is still zero.
In fact, the magnetic field generated by the wire forms concentric lines around the wire, so it is parallel to the plane of the coil. But the flux is equal to
where
Answer:
Among those two medium, light would travel faster in the one with a reflection angle of (when light enters from the air.)
Explanation:
Let denote the speed of light in the first medium. Let
denote the speed of light in the air. Assume that the light entered the boundary at an angle of
to the normal and exited with an angle of
. By Snell's Law, the sine of
and
would be proportional to the speed of light in the corresponding medium. In other words:
.
When light enters a boundary at the critical angle , total internal reflection would happen. It would appear as if the angle of refraction is now
. (in this case,
.)
Substitute this value into the Snell's Law equation:
.
Rearrange to obtain an expression for the speed of light in the first medium:
.
The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.
For ,
is monotonically increasing with respect to
. In other words, for
in that range, the value of
increases as the value of
increases.
Therefore, compared to the medium in this question with , the medium with the larger critical angle
would have a larger
. such that light would travel faster in that medium.
This assumes that the wave has velocity c (is light).
