Final answer:
The electrical power dissipated in a coil when the number of turns is halved and the radius is doubled would remain unchanged because the decrease due to halved turns is offset by an increase due to the squared factor from the doubled radius, assuming constant resistance.
Step-by-step explanation:
The question pertains to the effect on the electrical power dissipated in a coil when the number of its turns is halved and the radius of the wire is doubled while in a time-varying magnetic field. According to Faraday's law of electromagnetic induction, the induced emf is directly proportional to the rate of change of the magnetic flux, which in turn is proportional to the number of turns in the coil and the area it encloses. Halving the turns reduces the flux change by half. However, doubling the radius of the coil increases the area by a factor of four (since area is proportional to the square of the radius), resulting in the magnetic flux being doubled. Consequently, the overall change in the induced emf and the subsequent current would be the same, assuming that the resistance remains constant. Therefore, given the power dissipation (P) in a resistor is given by P = I2R, with the current (I) unchanged and resistance (R) unchanged, the power dissipated would remain unchanged.