Question

A rectangular coil of *N *turns and of length *a *and width *b *is rotated at frequency $f$* *in a uniform magnetic field $B$, as indicated in Figure . The coil is connected to co-rotating cylinders, against which metal brushes slide to make contact. (a) Show that the

emf induced in the coil is given (as a function of time *t*) by

$E=2πfNab sin2πft=E_{0}sin2πft.$

This is the principle of the commercial alternating-current generator. (b) What value of *Nab *gives an emf with $E_{0}=150V$ when the loop is rotated at 60.0 rev/s in a uniform magnetic field of 0.500 T?

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Solution

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(a) It should be emphasized that the result, given in terms of $sin(2πft)$, could as easily be given in terms of $cos(2πft)$ or even $cos(2πft+ϕ)$ where $ϕ$ is a phase constant . The angular position $θ$ of the rotating coil is measured from some reference line (or plane), and which line one chooses will affect whether the magnetic flux should be written as $BAcosθ,BAsinθ$ or $BAcos(θ+ϕ).$ Here our choice is such that $Φ_{B}=BAcosθ.$ since the coil is rotating steadily, $θ$ increases linearly with time. Thus, $θ=ωt$ (equivalent to $θ=2πft$) if $θ$ is understood to be in radians (and $ω$ would be the angular velocity). since the area of the rectangular coil is $A=ab$,

Faraday's law leads to

$ε=−Ndtd(BAcosθ) =−NBAdtdcos(2πft) =NBab2πfsin(2πft)$

which is the desired result, shown in the problem statement. The second way this is written $(a_{0}sin(2πf))$ is meant to emphasize that the voltage output is sinusoidal (in its time dependence) and has an amplitude of $ε_{0}=2πfNabB$

(b) We solve

$ε_{0}=150V=2πfNabB$

when $f=60.0$ rev/s and $B=0.500$

T. The three unknowns are $N,$ a, and $b$ which occur in a product; thus,

we obtain $Nab=0.796m_{2}$

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