Final answer:
The peak voltage induced in a rectangular conducting loop spinning at 7200 revolutions per minute in a 6 x 10^-6 T magnetic field is 2.26 mV.
Step-by-step explanation:
The question is asking to determine the peak voltage induced in a rectangular conducting loop that is spinning in a magnetic field. To solve this, we use Faraday's Law of Electromagnetic Induction, which states that the induced electromotive force (emf) in any closed circuit is equal to the time rate of change of the magnetic flux through the circuit. The peak voltage can be calculated using the formula:
V = NABωsin(ωt),
where:
- N is the number of turns (in this case, 1 since it is a single loop),
- A is the area of the loop,
- B is the magnetic field strength,
- ω is the angular velocity,
- t is the time which is variable.
Since the question gives us the dimensions of the loop, magnetic field strength, and the spinning rate, we can calculate the peak voltage by first converting the spinning rate from revolutions per minute to radians per second and then using the above formula.
The angular velocity (ω) in rad/s is given by: ω = 2π * (revolutions per minute / 60). For a spinning rate of 7200 rpm, this is: ω = 2 * π * (7200 / 60) = 2 * π * 120 = 240π rad/s.
The area (A) is 5 cm x 10 cm = 0.0005 m².
Substituting the given values into Faraday's Law, we get:
Peak Voltage (V_peak) = NABω = (1) * (0.0005 m²) * (6 x 10^-6 T) * (240π rad/s) = 0.0005 * 6 x 10^-6 * 240π = 2.26 x 10^-3 V or 2.26 mV.