Question

Uniform displacement-current density. The figure shows a circular region of radius R = 3.70 cm in which a displacement current is directed out of the page.The displacement current has a uniform density of magnitude J_d = 8.90 A/m². What is the magnitude of the magnetic field due to the displacement current at radial distances (a) 1.40 cmand (b) 4.50 cm?

(In nT)

Answer 1

given the radius of the circular cross section = 3.70cm

.The displacement current has a uniform density of magnitude Jd = 8.90 A/m2

. we know current density jd = i/A

area of circular section = 3.14* 0.037*0.037=0.00429m^2

now current i = jd *A

i= 8.90*0.00429

i = 0.0382

a)

now magnitude of the magnetic field B = u0i/2pir

B = 4pi*10 -7 * 0.0382/2pi*0.014

B = 5.465*10 -7 T

b) now magnitude of the magnetic field B = u0i/2pir

B = 4pi*10 -7 * 0.0382/2pi*0.045

B = 1.697*10 -7 T