An ac generator consists of 100 loops of wire, each of area $0.090 \mathrm {~m} ^ { 2 }$, and has a total resistance $12 \Omega$ The loops rotate about a diameter in a magnetic field of $0.50 \mathrm {~T}$ at a constant angular speed of 60 revolutions per second. Find the maximum induced emf in the generator.
A) $0.54 \mathrm { kV }$
B) $1.7 \mathrm { kV }$
C) $0.27 \mathrm { kV }$
D) $3.4 \mathrm { kV }$

Question

Quizplus · Accepted Answer

The maximum induced emf ($E_{max}$) in an AC generator can be calculated using the formula $E_{max} = NAB\omega$, where $N$ is the number of loops, $A$ is the area of each loop, $B$ is the magnetic field strength, and $\omega$ is the angular velocity in radians per second. Given $N = 100$, $A = 0.090 \, \text{m}^2$, $B = 0.50 \, \text{T}$, and the angular speed is 60 revolutions per second (which needs to be converted to radians per second by multiplying by $2\pi$, since there are $2\pi$ radians in a revolution), the angular velocity $\omega = 60 \times 2\pi = 120\pi \, \text{rad/s}$. Plugging these values into the formula gives $E_{max} = 100 \times 0.090 \times 0.50 \times 120\pi$, which calculates to approximately $1700 \, \text{V}$ or $1.7 \, \text{kV}$.

An Ac Generator Consists of 100 Loops of Wire, Each 0.090 m20.090 \mathrm {~m} ^ { 2 }0.090 m2

An Ac Generator Consists of 100 Loops of Wire, Each $0.090 \mathrm {~m} ^ { 2 }$