The simple loop is rotating in a uniform magnetic field shown in Figure 3-1 has the following characteristics: B =1.0 T to the right r = 0.1 m l = 0.3 m ωm = 377 rad/s (a) Calculate the voltage etot(t) induced in this rotating loop. (b) What is the frequency of the voltage produced in this loop? (c)

Question

The simple loop is rotating in a uniform magnetic field shown in Figure 3-1 has the following characteristics:

[latex]\mathbf{B}=1.0 \mathrm{~T} ext { to the right } \quad r=0.1 \mathrm{~m} \ l=0.3 \mathrm{~m} \quad \omega_m=377 \mathrm{\ rad} / \mathrm{s}[/latex]

(a) Calculate the voltage [latex]e_{\mathrm{tot}}(t)[/latex] induced in this rotating loop.
(b) What is the frequency of the voltage produced in this loop?
(c) Suppose that a 10 Ω resistor is connected as a load across the terminals of the loop. Calculate the current that would flow through the resistor.
(d) Calculate the magnitude and direction of the induced torque on the loop for the conditions in (c).
(e) Calculate the instantaneous and average electric power being generated by the loop for the conditions in (c).
(f) Calculate the mechanical power being consumed by the loop for the conditions in (c). How does this number compare to the amount of electric power being generated by the loop?

Accepted Answer

(a) The induced voltage on a simple rotating loop is given by

[latex]\begin{aligned}& e_{ ext {ind }}(t)=2 r \omega B l \sin \omega t \quad \quad \quad ext {(3-8)}\& e_{ ext {ind }}(t)=2(0.1 \mathrm{~m})(377 \mathrm{\ rad} / \mathrm{s})(1.0 \mathrm{~T})(0.3 \mathrm{~m}) \sin 377 t \& e_{ ext {ind }}(t)=22.6 \sin 377 t \mathrm{~V}\end{aligned}[/latex]

(b) The angular velocity of the voltage produced in the loop is 377 rad/s. Frequency is related to angular velocity by the equation [latex]\omega=2 \pi f[/latex], so

[latex]f=\frac{\omega}{2 \pi}=\frac{377}{2 \pi}=60 \mathrm{~Hz}[/latex]

(c) If a 10 Ω resistor is connected as a load across the terminals of the loop, the current flow would be:

[latex]i(t)=\frac{e_{ ext {ind }}}{R}=\frac{22.6 \sin 377 t \mathrm{~V}}{10 \ \Omega}=2.26 \sin 377 t \mathrm{~A}[/latex]

(d) The induced torque would be:

[latex]\begin{aligned} au_{ ext {ind }}(t) & =2 ext { rilB } \sin heta \quad \quad \quad ext {(3-17)}\ au_{ ext {ind }}(t) & =2(0.1 \mathrm{~m})(2.26 \sin \omega t \mathrm{~A})(0.3 \mathrm{~m})(1.0 \mathrm{~T}) \sin \omega t \ au_{ ext {ind }}(t) & =0.136 \sin ^2 377 t \mathrm{~N} \cdot \mathrm{m}, ext { counterclockwise }\end{aligned}[/latex]

(e) The instantaneous power generated by the loop is:

[latex]P(t)=e_{ ext {ind }} i=(22.6 \sin 377 t \mathrm{~V})(2.26 \sin 377 t \mathrm{~A})=51.1 \sin ^2 377 t \mathrm{~W}[/latex]

The average power generated by the loop is

[latex]P_{ ext {ave }}=\frac{1}{T} \int_T 51.1 \sin ^2 \omega t \ d t=25.55 \mathrm{~W}[/latex]

(f) The mechanical power being consumed by the loop is:

[latex]P= au_{ ext {ind }} \omega=\left(0.136 \sin ^2 377 t \mathrm{~N}⋅\mathrm{m} ight)(377 \mathrm{\ rad} / \mathrm{s})=51.3 \sin ^2 \omega t \mathrm{~W}[/latex]

Note that the amount of mechanical power consumed by the loop is equal to the amount of electrical power created by the loop (within roundoff error). This machine is acting as a generator, converting mechanical power into electrical power.

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