Question

module 5 practice final exam - attempt 1
item 5
a 590 - loop circular armature coil with a diameter of 9.0 cm rotates at 190 rev/s in a uniform magnetic field of strength 0.70 t.
part a
what is the rms voltage output of the generator?
express your answer to two significant figures and include the appropriate units.
v_{rms} =
value
units
submit request answer
part b
what would you do to the rotation frequency in order to double the rms voltage output?
to double the output voltage, you must triple the rotation frequency
you must double the rotation frequency

Explanation:

Step1: Find the area of the coil

The diameter \( d = 9.0\ \text{cm}=0.09\ \text{m} \), so the radius \( r=\frac{d}{2}=0.045\ \text{m} \). The area of a circle is \( A = \pi r^{2} \), so \( A=\pi(0.045)^{2}\approx 6.36\times 10^{-3}\ \text{m}^2 \).

Step2: Find the angular frequency

The rotation frequency \( f = 190\ \text{rev/s} \), and angular frequency \( \omega = 2\pi f \), so \( \omega=2\pi\times190\approx 1194\ \text{rad/s} \).

Step3: Find the maximum emf

The formula for the maximum emf of a generator is \( \mathcal{E}_{\text{max}}=NAB\omega \), where \( N = 590 \), \( A = 6.36\times 10^{-3}\ \text{m}^2 \), \( B = 0.70\ \text{T} \), \( \omega = 1194\ \text{rad/s} \). Plugging in the values: \( \mathcal{E}_{\text{max}}=590\times6.36\times 10^{-3}\times0.70\times1194\approx 590\times6.36\times 10^{-3}\times835.8\approx 590\times5.31\approx 3133\ \text{V} \).

Step4: Find the rms voltage

The rms voltage \( V_{\text{rms}}=\frac{\mathcal{E}_{\text{max}}}{\sqrt{2}} \), so \( V_{\text{rms}}=\frac{3133}{\sqrt{2}}\approx 2215\ \text{V}\approx 2.2\times 10^{3}\ \text{V} \) (to two significant figures).

Answer:

\( V_{\text{rms}}=\boxed{2.2\times 10^{3}\ \text{V}} \) (or \( 2.2\ \text{kV} \))