An ac generator with a total resistance of 12 Ω contains 100 flat loops of wire,each with an area of 0.090 m².The loops rotate at 60 rev/s in a $\vec { B }$ field of magnitude 0.50 T.What is the maximum possible induced current? A)23 A B)46 A C)0.14 kA D)0.28 kA

Question

Quizplus · Accepted Answer

The maximum possible induced current can be found using the formula:

\begin{align*}
I_{max}&=\frac{\omega B A N}{R}\
&=\frac{2 \pi f B A N}{R}
\end{align*}

where $\omega$ is the angular velocity, $f$ is the frequency, $B$ is the magnetic field strength, $A$ is the area of each loop, $N$ is the number of loops, and $R$ is the total resistance.

Substituting the given values:

\begin{align*}
I_{max}&=\frac{2 \pi (60	ext{ rev/s}) (0.50	ext{ T}) (0.090	ext{ m}^2) (100)}{12	ext{ }\Omega}\
&= 0.14	ext{ kA}
\end{align*}

Therefore, the answer is choice C.

An Ac Generator with a Total Resistance of 12 Ω B⃗\vec { B }B

An Ac Generator with a Total Resistance of 12 Ω $\vec { B }$