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 180-turn coil that has an area per turn of 0.94 m2. What should be the magnitude of the magnetic field in which the coil rotates?

Answer 1

Answer:

0.082 T

Explanation:

Given:

Frequency of the generator (f) = 60.0 Hz

Maximum emf of the generator (E) = 5200 V

Number of turns (N) = 180

Area per turn (A) = 0.94 m²

Magnetic field magnitude (B) = ?

We know that, the angular frequency of the generator is given as:

$\omega=2\pi f$

Plug in the value of 'f' and solve for 'ω'. This gives,

$\omega=2\pi\times 60.0=120\pi\ rad/s$

Now, the maximum emf of the generator is given by the formula:

$E=NAB\omega$

Rewriting in terms of magnetic field, 'B', we get:

$B=\frac{E}{NA \omega}$

Plug in the given values and solve for 'B'. This gives,

$B=\frac{5200\ V}{180\times 0.94\ m^2\times 120\pi\ rad/s}\\\\B=\frac{5200\ V}{63786.897}\\\\B=0.082\ T$

Therefore, the magnitude of the magnetic field in which the coil rotates is 0.082 T.