Question

[G 7.24) An alternating current Io cos (wt) (amplitude 0.5 A, frequency 60 Hz) flows down a straight wire, which

runs along the aris of a toroidal coil with rectangular cross section (inner radius 1 cm, outer radius 2 cm, height 1

cm, 1000 turns). The coil is connected to a 500 ohm resistor.

(a) In the quasistatic approximation, what emf is induced in the toroid? Find the current, I, (t), in the resistor.

b) Calculate the back emf in the coil, due to the curent I, (t). What is the ratio of the amplitudes of this back emf

and the "direct" emf in ( a ) ?

Answer 1

a)

$\Phi=N\int _a^b\int_0^h\frac{\mu _0 I(t)}{2\pi s}dsdz=\frac{\mu _0 N I(t)}{2\pi }h \text{ln}\frac ba,$

$\varepsilon=-\frac{d \Phi}{dt}=-\frac{\mu _0 N }{2\pi }h \text{ln}\frac ba \frac{dI}{dt}=\frac{\mu _0 N h \omega I_0 }{2\pi } \text{sin }\omega t\text{ln}\frac ba=2.61\cdot 10^{-4}~\text{V},$

$I_R=\frac{\varepsilon}{R}=\frac{\mu _0 N h \omega I_0 }{2\pi R }\text{ sin} \omega t\text{ln}\frac ba=5.23\cdot 10^{-7}~\text{A},$

b)

$\varepsilon _b=-L\frac{dI_R}{dt}=-(\frac{\mu _0 N^2 h }{2\pi }\text{ln}\frac ba )(-\frac{\mu _0 N h \omega^2 I_0 }{2\pi R } \text{cos}\omega t\text{ln}\frac ba)=\frac{\mu _0^2 N^3 h ^2\omega^2 I_0 }{4\pi^2 R } \text{cos}\omega t\text{ln}^2\frac ba=2.73\cdot 10^{-7}~\text{V},$ $\frac{\varepsilon_b}{\varepsilon}=\frac{\mu _0 N^2 h \omega }{2\pi R } \text{ln}\frac ba=1.05\cdot 10^{-3}.$