A long, straight, copper wire with a circular cross-sectional area of 2.1 mm 2 (

Question

A long, straight, copper wire with a circular cross-sectional area of

2.1 mm2

(a) What is the uniform electric field in the material?
V/m

(b) If the current is changing at the rate of 3500 A/s, at what rate is the electric field in the material changing?
V/m·s

(c) What is the displacement current density in the material in part (b)? (Hint: Since K for copper is very close to 1, use = 0.)
A/m2

(d) If the current is changing as in part (b), what is the magnitude of the magnetic field 6.0 cm from the center of the wire? Note that both the conduction current and the displacement current should be included in the calculation of B.
T

Explanation / Answer

A)

Current density in the wire, J = I/A

= 17/(2.1*10^-6)

= 8.095*10^6 A/m^2

Alternate equation for Ohm's law

J = sigma*E (here J is current density,sigma is condtuctivity and E is electric filed)

or J = E/rho (here rho is resistivity)

==> E = J*rho

= 8.095*10^6*2*10^-8

= 0.162 V/m

b) E = I*rho/A

so, dE/dt = (rho/A)*dI/dt

= (2*10^-8/(2.1*10^-6))3500

= 33.33 V/(m.s)

c) we know dispalcement current, Id = A*epsilon*dE/dt

= 2.1*10^-6*8.854*10^-12*3500

= 6.5*10^-14 T

d) Apply, B = mue*Id/(2*pi*r)

= 4*pi*10^-7*6.5*10^-14/(2*pi*0.06)

= 2.17*10^-19 T

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