A moving electron passes near the nucleus of a gold atom, which contains 79 protons and 118 neutrons. At a particular moment the
1 answer:
Answer:
Explanation:
Given data
Distance r=7.5×10⁻⁹m
Charge of electron -e= -1.6×10⁻¹⁹C
Charge of proton e=1.6×10⁻¹⁹C
To find
Electric force F
Solution
From Coulombs law we know that:
q₁ is charge of electron
q₂ is the charge of gold nucleus which contains 79 positively charge protons and 118 neutral neutrons.
The Charge of single proton e=1.6×10⁻¹⁹C
79 proton charge q₂=79×1.6×10⁻¹⁹=1.264×10⁻¹⁷C
So
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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