Question

Electricity and Magnetism: Will rate high for correct solutions!

The figure below shows a cross-section of a long (infinite for our purposes) solenoidal system consisting of two coaxial solenoids, of radii a=3.3 cm and b=22.9 cm. These solenoids have the same number n=10 cm-1 of the turns of the wire per unit length. The wire is common and fed by the time t-dependent current I(t)=Io.sin(29f.t) but in the opposite directions as illustrated in the figure. Here the current amplitude Io=24.5 A and frequency f=120 Hz. In this problem, we are interested in the electric field induced in the space by this system. With the harmonically varying source current, the induced electric field will also be changing periodically with the same frequency. You are asked to examine the amplitude Eo() of the induced electric field at different distances r from the axis of the system. What is Eo(r) for r=r1=1.9 cm?: Eolri)= V/m. What is Eo(r) for r=r2=14.2 cm?: Eo(r2)= V/m. What is Eo(r) for r=r3=33.8 cm?: Eo(r3)= V/m.

Answer 1

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nr 2 un m Part a: For r = ri = 1.9 cm <a dф d ſ Ede (Bara) dt dt dB E(21) dt rdB E 2 dt B = Moln dB di = un 1. (20f) cos(2nft) dt dt Then: E = -Honlaf r cos(2nft) The amplitude is: E. = u.nl.nfrı = (4110-7)(1000) (24.5)(120)(0.019) V E. = 0.2205 Part b: For r = r2 = 14.2 cm a<r<b Then: dB E(21) dB di dt = un = un 1. (2f) cos(27ft) dt E2nr) = -M.nl. (20f) cos(2nft) (r2 – a?) Monl.fr(r2 – a) cos(2nft) The amplitude is: H.nl.fr(r2 – a) E. Using the given data: (4110-7)(1000) (24.5)(120)(0.1422 – 0.0332) E. 1.46 0.142 = dt (r² – a²) E = r r V m Part c: V E. u.nl.fr(b2 – a) (4110-7)(1000) (24.5)(120) (0.2292 – 0.0332) = 1.76– 0.338 = r m