91Ó°ÊÓ

As per Faraday's law of electromagnetic induction, induced emf (ε) is equal to the time derivative of the magnetic flux through the loop. The loop in this case consists of the semicircular wire. To calculate the induced emf in the wire, we first need to know the magnetic flux and find the derivative of it concerning time. The magnetic flux (Ф) through the loop can be given by: Ф = \(B \times A\), where A is the area of the semicircle The area of the semicircular loop is (1/2) * (π * r^2). So, we have: Ф = \(B \times \frac{1}{2} \times \pi r^{2}\) Now, we find the time derivative of the magnetic flux, dФ/dt: dФ/dt = \(\frac{d}{dt}(B \times \frac{1}{2} \times \pi r^{2})\) As 'r' and 'B' are constants, taking the derivative concerning time, we get: dФ/dt = \(B \pi r^{2} \times \frac{d}{dt}( \frac{1}{2})\) Now we can find the induced emf (ε), which is equal to the time derivative of the magnetic flux. ε = - dФ/dt

The time derivative of 1/2 is related to the rate of change of angle of rotation concerning time: \(\frac{1}{2} \times \frac{dθ}{dt}\) = \(\frac{1}{2} ω\)

Substituting the expression for the rate of change of angle in the equation of induced emf, we get: ε = - B π r^2 * (\(\frac{1}{2} ω\)) = -\(\frac{B \pi r^{2} \omega}{2}\) The negative sign represents the reverse movement of the charges and actual flow.

Now, we can find the mean power generated in the circuit using the formula: P = \(\frac{ε^{2}}{R}\) Substitute the induced emf value: Mean power, P = \(\frac{(\frac{-B \pi r^{2} \omega}{2})^{2}}{R}\) P = \(\frac{\left(B \pi r^{2} \omega\right)^{2}}{4 R}\) Taking the mean power per period, we have to divide the power by 2: Mean power per period = \(\frac{\left(B \pi r^{2} \omega\right)^{2}}{8 R}\) Comparing the obtained expression with the options in the exercise, we find that the correct answer is: (D) \(\frac{\left(B \pi r^{2} \omega\right)^{2}}{8 R}\)