(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.
Answer:
Generally, magnets are attracted to objects that are made of the metals iron, nickel, or cobalt. These materials are called ferromagnetic materials. ... When all or most of the domains are aligned in the same direction, the whole object becomes magnetized in that direction and becomes a magnet.
Explanation:
Let us take east and north as the positive x and y-axes should the motion be plotted in a cartesian plane. Thus, the x value is 45 miles and the y value is 20. The tangent of an angle is equal to the ratio of y to x.
tanθ = y / x
Substituting,
tanθ = 20/45 = 0.44
The value of θ is 23.96°.
Answer: The electric field is: a) r<a , E0=; b) a<r<b E=ρ (r-a)/εo;
c) r>b E=ρ b (b-a)/r*εo
Explanation: In order to solve this problem we have to use the Gaussian law in diffrengios regions.
As we know,
∫E.dr= Qinside/εo
For r<a --->Qinside=0 then E=0
for a<r<b er have
E*2π*r*L= Q inside/εo in this case Qinside= ρ.Vol=ρ*2*π*r*(r-a)*L
E*2π*r*L =ρ*2*π*r* (r-a)*L/εo
E=ρ*(r-a)/εo
Finally for r>b
E*2π*r*L =ρ*2*π*b* (b-a)*L/εo
E=ρ*b* (b-a)*/r*εo
Answer:
Infra and Red
Explanation:
While many elemental spectral lines are visible, almost all molecular lines lie in the infra portion of the spectrum, since they are at much lower energy. while many elemental spectral lines are visible, almost all molecular lines lie in the red portion of the spectrum, since they are at much lower energy.