Nonuniform displacement-current density. The figure below shows a circular regio

Question

Nonuniform displacement-current density. The figure below shows a circular region of radius R = 3.30 cm in which a displacement current is directed out of the page. The magnitude of the density of this displacement current is given by Jd = (3.80 A/m2)(l - r/R), where r is the radial distance (r R). What is the magnitude of the magnetic field due to the displacement current at a radial distance of 2.00 cm? 17.86666 times nT What is the magnitude of the magnetic field due to the displacement current at a radial distance of 5.00 cm? nT

Explanation / Answer

These questions are really insane. I teach physics to pre-medstudents and to engineering students. These would be consideredsome of the medium-level problems for the engineering students. Ifthat's what you're heading for, then these are ok. If not... yourprof is a nutcase. You have my condolences. . Anyway... the problem itself isn't too complicated, but it'sjust so unnecessary. . The idea is that the B field is givenby     B = k I /r       where I is thecurrent "enclosed" by a loop of the given radius. . To find that current, you have to integrate the currentdensity: .    total current = integral of ( currentdensity * dA0 = integral of (current density *2r dr) = .             integral of ( 3.80 (1 - r/R) 2r dr )  =      7.60   integral of( r - r2 / R )dr   = .             =   7.60 [ (1/2) r2  -   (1/3) r3 / R ] . So now... part a: .    r = 2.00 cm    R = 3.30cm       .       total current = I =7.60 [ (1/2) * 0.022 - (1/3) *0.023 / 0.033 ] = 0.0028458 amps . B field = k I / r = 2 x10-7 * 0.0028458 / 0.02   = 0.28458 x10-7 =   28.46 nT . For part b, at a distance of 5.00 cm you are beyond thedisk. You must use r = 3.30 cm to evaluate thecurrent: .      I = 7.60 [ (1/2) *0.03302    -   (1/3) *0.0333 / 0.033 ]   =   0.0043335amps. . and then . B field = k I / r = 2 x10-7 * 0.0043335 / 0.05   = 0.17334 x10-7=   17.33nT

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