Asilver wire has resistivity rho = 1.94 × 10-8 ω·m and a cross-sectional area of 7.00 mm2. the current in the wire is uniform and changing at the rate of 2400 a/s when the current is 180 a.

(a) what is the magnitude of the (uniform) electric field in the wire when the current in the wire is 180 a?
(b) what is the displacement current in the wire at that time?
(c) what is the ratio of the magnitude of the magnetic field due to the displacement current to that due to the current at a distance r from the wire?

(a). The magnitude of the electric field in the wire is 0.498 V/m.

(b). The displacement current in the wire at that time is

(c). The ratio of the magnitude of the magnetic field due to the displacement current to that due to the current at a distance r from the wire is

Explanation:

Given that,

Resistivity

Area of cross section area A =7.00 mm²

Rate = 2400 A/s

Current = 180 A

(a). We need to calculate the magnitude of the electric field in the wire

Using for of electric field

Put the value into the formula

(b). We need to calculate the displacement current in the wire

Using formula of displacement current

Put the value into the formula

(c). We need to calculate the ratio of the magnitude of the magnetic field due to the displacement current to that due to the current at a distance r from the wire

Using formula of magnetic field

Put the value into the formula

Hence, (a). The magnitude of the electric field in the wire is 0.498 V/m.

(b). The displacement current in the wire at that time is

(c). The ratio of the magnitude of the magnetic field due to the displacement current to that due to the current at a distance r from the wire is

no. dot product   =0 when vecs at right angles