A small circular loop of 5 mm radius is placed 1 m away from a 60-Hz power line. The voltage induced on this loop is measured at
0.6 µV. What is the current on the power line
1 answer:
Answer:
i = 101.4A
Explanation:
The steps to the solution can be found in the attachment below.
We have been given the frequency f = 60Hz. From this we can calculate the angular frequency of the power line.
It is assumed that sinwt = –1 in order to calculate the time varying current. Although the magnitude of the current is large, consideration should also be given to the distance between the coil and the power line. The induced emf is small considering the area of the coil which is 7.85×10-⁵m².
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(1) The position around equilibrium of an object in simple harmonic motion is described by
where
A is the amplitude of the motion
is the angular frequency.
The velocity is the derivative of the position:
where
is the maximum velocity of the object.
The acceleration is the derivative of the velocity:
where
is the maximum acceleration of the object.
We know from the problem both maximum velocity and maximum acceleration:
From the first equation, we get
(1)
and if we substitute this into the second equation, we find the angular fequency
while the amplitude is (using (1)):
(b) We found in the previous step that the angular frequency of the motion is
But the angular frequency is related to the period by
and so, the period is
where
A is the amplitude of the motion
The velocity is the derivative of the position:
where
The acceleration is the derivative of the velocity:
where
We know from the problem both maximum velocity and maximum acceleration:
From the first equation, we get
and if we substitute this into the second equation, we find the angular fequency
while the amplitude is (using (1)):
(b) We found in the previous step that the angular frequency of the motion is
But the angular frequency is related to the period by
and so, the period is