Question

You need to design a 60.0-Hz ac generator that has a maximum emf of 5200 V. The generator is to contain a 130-turn coil that has an area per turn of 0.82 m2. What should be the magnitude of the magnetic field in which the coil rotates

Answer 1

Answer:

B =  0.129 T

Explanation:

Given,

frequency, f = 60 Hz

maximum  emf = 5200 V

Number of turns, N = 130

Area per turn = 0.82 m²

We know,

ω = 2 π f

ω = 2 π x 60 = 376.99 rad/s

now, Magnetic field calculation

$B =\dfrac{\epsilon_{max}}{NA\omega}$

$B =\dfrac{5200}{130\times 0.82\times 376.99}$

B =  0.129 T

Hence, the magnetic field is equal to B =  0.129 T