A circular wire loop lies inside a region of space containing a magnetic field. The direction of the magnetic field is out of th
1 answer:
You might be interested in
<h3>Question:</h3>
A 2.0-cm length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnetic field at the point x = 5.0m on the x-axis.
<h3>Answer:</h3>
1.6nT [in the negative z direction]
<h2>Explanation:</h2>
The magnetic field, B, due to a distance of finite value b, is given by;
B = (μ₀IL) / (4πb) -----------(i)
Where;
I = current on the wire
L = length of the wire
μ₀ = magnetic constant = 4π × 10⁻⁷ H/m
From the question,
I = 20A
L = 2.0cm = 0.02m
b = 5.0m
Substitute the necessary values into equation (i)
B = (4π × 10⁻⁷ x 20 x 0.02) / (4π x 5.0 )
B = (10⁻⁷ x 20 x 0.02) / (5.0 )
B = (10⁻⁷ x 20 x 0.02) / (5.0 )
B = (10⁻⁷ x 20 x 0.02) / (25.0)
B = 1.6 x 10⁻⁹T
B = 1.6nT
Therefore, the magnetic field at the point x = 5.0m on the x-axis is 1.6nT.
PS: Since the current is directed in the positive y direction, from the right hand rule, the magnetic field is directed in the negative z-direction.
4) The correct answer is:
<span>B. An electron will be emitted in the second experiment, but it cannot be determined whether it will reach the second plate.
In fact, violet light has higher frequency than green light. This means that photons of violet light carry more energy than photons of green light (remember that the energy of a photon is proportional to its frequency: </span>
)<span>, so when they hit the surface of the metal, more energy is transferred to the electrons. The electron was already emitted with green light, so it must be emitted also with violet light, given the more energy transferred. The electron will also have more kinetic energy when hit by violet light, however, we cannot determine if it will reach the second plate, since we don't know how much energy has been used to extract the electron from the metal (in fact, we don't know the work function of the metal, i.e. the energy needed to extract the electron)
3) The correct answer is
</span><span>A. Violet light will cause electrons to be emitted at greater velocities than those removed by green light.
In fact, </span>violet light has higher frequency than green light. This means that photons of violet light carry more energy than photons of green light (remember that the energy of a photon is proportional to its frequency:), so when they hit the surface of the metal, more energy is transferred to the electrons. Therefore, the emitted electrons will have on average greater energy (and so, greater velocity) than those removed by green light.
<span>B. An electron will be emitted in the second experiment, but it cannot be determined whether it will reach the second plate.
In fact, violet light has higher frequency than green light. This means that photons of violet light carry more energy than photons of green light (remember that the energy of a photon is proportional to its frequency: </span>
3) The correct answer is
</span><span>A. Violet light will cause electrons to be emitted at greater velocities than those removed by green light.
In fact, </span>violet light has higher frequency than green light. This means that photons of violet light carry more energy than photons of green light (remember that the energy of a photon is proportional to its frequency: