Submit Answer Save Progress Practice Another Version Notes Ask Your Teacher 0-12

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Submit Answer Save Progress Practice Another Version Notes Ask Your Teacher 0-12 points HRW9 32.P021 Nonuniform displacement-current density. The figure below shows a circular region of radius R -3.20 cm in which a displacement current is directed out of the page. The magnitude of the density of this displacement current is given by J (2.60 A/m2)(1 - r/R), where r is the radial distance (r s R). (a) what is the magnitude of the magnetic field due to the displacement current at a radial distance of 2.00 cm? (b) What is the magnitude of the magnetic field due to the displacement current at a radial distance of 5.00 cm? nT Read the eBook Section 32-4 Displacement Current 8. 4/4 points1 Previous Answers HRW9 32. P.O23.ssm. Notes Ask Your Teacher

Explanation / Answer

R = 3.2 cm = 3.2*10^-2 m

Jd = 2.6(1 - r/R)

a. at r = 2 cm

from ampere's law

B*2*pi*r = ien*mu

but

to find ien

di = 2*pi*r*dr*J(r) = 2.6(1 - r/R)*2*pi*rdr

integrating from r = 0 to r = r

i = 2.6*2*pi(r^2/2 - r^3/3R)

hence

B = 2.6*2*pi(r^2/2 - r^3/3R)*mu/2*pi*r

B = 10.4*pi(0.02/2 - 0.02^2/3R)*10^-7

B = 0.190589*10^-7 T

b. for r = 5 cm

i= 2.6*2*pi(R^2/2 - R^3/3R) = 2.6*pi(R^2)/3 = 0.002788 A

hence

B*2*pi*0.05 = 4*pi*10^-7*i

B = 0.11152235 *10^-7 T

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