Question

A length of 4.000 m of wire is to be used to detect a magnetic field. The wire is made into a single square loop and rotated at a rate of 400 cycles per second. If the magnetic field is 0.02000 T, what is the magnitude of the average emf that can be generated in the first quarter cycle, provided the loop is initially oriented in a plane perpendicular to the magnetic field?

a) 2.0 V
b) 4.0 V
c) 8.0 V
d) 16.0 V

Answer 1

The magnitude of the average emf that can be generated in the first quarter cycle is 8.0 V.

Step-by-step explanation:

To calculate the magnitude of the average emf generated in the first quarter cycle, we can use the formula: emf = NABωsinθ, where N is the number of loops, A is the area of one loop, B is the magnetic field strength, ω is the angular frequency, and θ is the angle between the normal of the loop and the magnetic field.

In this case, we have a single square loop with a side length of 4.000 m, so the area of one loop is A = (4.000 m)^2 = 16.000 m^2.

The angular frequency ω can be calculated using the formula ω = 2πf, where f is the frequency in cycles per second.

Given that the frequency is 400 cycles per second, ω = 2π * 400 rad/s. The angle θ is 90 degrees since the loop is initially perpendicular to the magnetic field.

Finally, the average emf can be calculated using emf = NABωsinθ. Plugging in the values, we have emf = (1)(16.000 m^2)(0.02000 T)(2π * 400 rad/s)(sin 90°) = 8.000 V.

Therefore, the magnitude of the average emf that can be generated in the first quarter cycle is 8.0 V.