Answer:
(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

Step-by-step 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
