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
$E = 2 π f N a b sin 2 π f t = E_{0} sin 2 π f t .$
This is the principle of the commercial alternating-current generator. (b) What value of Nab gives an emf with $E_{0} = 150 V$ when the loop is rotated at 60.0 rev/s in a uniform magnetic field of 0.500 T?

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Answer 1

(a) It should be emphasized that the result, given in terms of $sin (2 π f t)$ , could as easily be given in terms of $cos (2 π f t)$ or even $cos (2 π f t + ϕ)$ 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 $B A cos θ, B A sin θ$ or $B A cos (θ + ϕ) .$ Here our choice is such that $Φ_{B} = B A cos θ .$ since the coil is rotating steadily, $θ$ increases linearly with time. Thus, $θ = ω t$ (equivalent to $θ = 2 π f t$ ) if $θ$ is understood to be in radians (and $ω$ would be the angular velocity). since the area of the rectangular coil is $A = a b$ ,
Faraday's law leads to
$ε = - N \frac{d ( B A cos θ )}{d t} = - N B A \frac{d cos ( 2 π f t )}{d t} = N B a b 2 π f sin (2 π f t)$
which is the desired result, shown in the problem statement. The second way this is written $(a_{0} sin (2 π f))$ is meant to emphasize that the voltage output is sinusoidal (in its time dependence) and has an amplitude of $ε_{0} = 2 π f N a b B$
(b) We solve
$ε_{0} = 150 V = 2 π f N a b B$
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 $N a b = 0.796 m^{2}$

Answer 2

(a) It should be emphasized that the result, given in terms of $sin (2 π f t)$ , could as easily be given in terms of $cos (2 π f t)$ or even $cos (2 π f t + ϕ)$ 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 $B A cos θ, B A sin θ$ or $B A cos (θ + ϕ) .$ Here our choice is such that $Φ_{B} = B A cos θ .$ since the coil is rotating steadily, $θ$ increases linearly with time. Thus, $θ = ω t$ (equivalent to $θ = 2 π f t$ ) if $θ$ is understood to be in radians (and $ω$ would be the angular velocity). since the area of the rectangular coil is $A = a b$ ,
Faraday's law leads to

$ε = - N \frac{d ( B A cos θ )}{d t} = - N B A \frac{d cos ( 2 π f t )}{d t} = N B a b 2 π f sin (2 π f t)$

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

(b) We solve

$ε_{0} = 150 V = 2 π f N a b B$

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 $N a b = 0.796 m^{2}$

A planar loops of wire rotates in a uniform magnetic field. Initially, at $t = 0$ , the plane of the loop is perpendicular to the magnetic field. If it rotates with a period of $10 s$ about an axis in its plane then the magnitude of induced emf will be maximum and minimum, respectively at :

The phase difference between induced emf and magnetic flux linked with a coil rotating in a uniform magnetic field is:

A square coil ACDE with its plane vertical is released from rest in a horizontal uniform magnetic field $B$ of length 2 L The acceleration of the coil is

Revise with Concepts

Gauss Law for Magnetism

Faraday Law Lenz's Law - Problem L1

A planar loops of wire rotates in a uniform magnetic field. Initially, at t=0, the plane of the loop is perpendicular to the magnetic field. If it rotates with a period of 10s about an axis in its plane then the magnitude of induced emf will be maximum and minimum, respectively at :