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.
You might be interested in
(a) The frequency of water wave is 2 Hz.
(b) The wave speed of the water wave is 3.6 m/s.
<u>Explanation:</u>
(a) It is known that completion of one complete wave in 1 second is defined as frequency of 1 HZ. So here there are 120 waves crossing the boat in 1 minute. So the frequency of the water wave will be
As the time is 1 minute which is equal to 60 seconds and the number of waves is given as 120 then the frequency of the water wave is
So the frequency of water wave is 2 Hz.
(b) Then if the wavelength of the water wave is 1.8 m with a frequency of 2 Hz, then speed of the wave can be determined as the product of wavelength with frequency.
So Speed = Frequency × Wavelength
Speed = 2 × 1.8 = 3.6 m/s.
So the speed of the water wave is 3.6 m/s.
Explanation:
It is given that,
Electric field,
We need to find the change in the electric potential energy of the proton-field system when the proton travels to x = 2.5 m
From the conservation of energy, the loss in potential energy is equal to the gain in kinetic energy and kinetic energy is is equal to the work done.
So, the change in electric potential energy of the proton field system is . Hence, this is the required solution.