Question

A simple generator is used to generate a peak output voltage of 25.0 V . The square armature consists of windings that are 7.0 cm on a side and rotates in a field of 0.490 T at a rate of 60.0 rev/s . Part A How many loops of wire should be wound on the square armature

Answer 1

Answer:

The 28 loops wound on the square armature

Explanation:

Peak output voltage $\epsilon _{peak} = 25$ V

Area of square armature $A = (7 \times 10^{-2} )^{2} = 49 \times 10^{-4}$

Magnetic field $B = 0.490$ T

Angular frequency $\omega = 2\pi f = 2 \pi \times 60 = 120\pi$

According to the law of electromagnetic induction,

     $\epsilon _{peak} = NBA \omega$

Where $N =$ number of loops of wire.

  $N = \frac{25}{49 \times 10^{-4} \times 0.49 \times 120\pi }$

  $N = 27.6$ ≅ 28

Thus, 28 loops of wire should be wound on the square armature.