Question

The figure below shows a cross-section of a long (infinite for our purposes) solenoidal system consisting of two coaxial solenoids, of radii a=4.3 cm and b=28.1 cm. These solenoids have the same number n=20 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(21f.t) but in the opposite directions as illustrated in the figure. Here the current amplitude lo=20.5 A and frequency f=148 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 Eolr) of the induced electric field at different distances r from the axis of the system. b а What is Eo(r) for r=r1=1.3 cm?: Eolr1)= V/m. What is Eo(r) for r=r2=23.3 cm?: Eolr2)= V/m. What is Eo(r) for r=r3=49.4 cm?: Eo(r3)= V/m.

Answer 1

Solution:- Ilt)= Io. Sin (21 ft) 4.3 cm. 28. 1c.m. Io = 20.5 A 148 HZ 0) Now, Eolo) for rar = 1.3c.m. DEM V dB (A) dt S E (219) s d - Molri 21Ta2 (TTM, 2) [ra] dt .. Eo (1) di 1 Mori 21tr 2ta 2 (Tr, 2) dt : Ee(n) Mo. ² . 1. (20f) 43 (217ft) 2 it 2a2 2 Mo. I Elr) CI Io (244) Cos (20 ft) ZIT 2a2

r=83 if we 1 by the Electric field for E(21183) dB (A) dt : beradses E (13) = 0 Because, consider a circle outside the Solenoid (876). Total current enclosed circle is I -I -0. So, Magnetic field is zero. In the region, acreb lieg g = 72 = 23.5 c.m., total flux d 21 . Electric field at raty a d dt Amplitude of Electric field, Parg Eo (3) uo (2014) en (ba) 2 IT Ed (3) (2x107) (201 x 148) en ( 286 ) 3.491 810-4 V/m Mo I(t) en (ba)