Complete Question
The complete question is shown on the first uploaded image
Answer:
a) 58 Hz b) 7.73 mV
Explanation:
From the complete question we are given that
Radius of the semi circle , a = 3.0 cm = (3.0)(1 × [tex]10^{-2}[/tex] m/1 cm)
=3.0 × [tex]10^{-2}[/tex] m
Angular speed ,w = 58 rev/s
= (58 rev/s)(2π rad /1 rev)
= 364.4 rad/s
Magnetic field, B = 15 mT
= (15 mT)(1 × [tex]10^{-3}[/tex] T/ 1 mT)
= 15 × [tex]10^{-3}[/tex] T
a) f = 58 rev/s
= 58 Hz
b) The Induced emf, ∈ = ωBπ[tex]a^{2}[/tex] /2
= [tex] ((364.4 rad/s)(15 × [tex]10^{-3}[/tex] T)(π)(3.0 × [tex]10^{-2}[/tex] m)^2)÷2[/tex]
= 0.007728 V
= 7.73 mV
The frequency of the loop is 58 Hz and the amplitude of the emf induced in the loop is 7.73 mV and this can be determined by using the given data.
Given :
A stiff wire bent into a semicircle of radius 'a' is rotated with a frequency f in a uniform magnetic field.
According to the data, the angular speed is 58 rev/sec that is, 364.4 rad/sec, the magnetic field is 15 mT that is, 15 [tex]\times[/tex] [tex]10^{-3}[/tex] T, and the radius 'a' is 3 cm that is, 3 [tex]\times[/tex] [tex]10^{-2}[/tex] m.
a) The frequency is 58 rev/sec that is 58 Hz.
b) The amplitude of the emf induced in the loop can be calculated as:
[tex]\rm \epsilon = \dfrac{\omega B \pi a^2}{2}[/tex]
Now, substitute the values of the known terms in the above formula.
[tex]\epsilon = \dfrac{364.4\times 15\times10^{-3}\times \pi \times (3\times 10^{-2})^2}{2}[/tex]
Further, simplify the above expression.
[tex]\rm \epsilon = 0.007728\;V[/tex]
[tex]\rm \epsilon = 7.73\;mV[/tex]
For more information, refer to the link given below:
https://brainly.com/question/4393505
