A long, straight, copper wire with a circular cross-sectional are of 1.5mmm^2 carries a current of 3 A
.
a).
E = I x rho/ A
= 3 x 2 x 10-8/ (1.5 x 10-6)
= 0.04V/m
b).
dE/dt= rho x dI/dt / A
= 8.8 x 105 V/m.s
c).
D= e0E
= 8.8 x 10-12 x 8.8 x 105
= 7.744 x 10-6 A/m2
d). Now,
id=(jd)A
id=(7.744E-6)*(1.5E-6)=11.6E-12
And generalized version of Amepre's law states
Line integral of B*dl=mu-0(ic+id)
ic=(dE/dt)*(A)*(e-0)=11.6E-12
So the Bc part is Bc=(ic/(2*pi*.06m))
= 3.08E-11
Bd= 7.96 (Use 3A as id for finding Bd)
e). not significant clear from from above.
A long, straight, copper wire with a circular cross-sectional are of 1.5mmm^2 carries a current of...
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