answer also given here...i just want you to rewrite use ur words

-25.59 . A 3.00-m length of copper wire at 20'C has a 1.20-m- long section with diameter 1.60 mm and a 1.80-m-long section with diameter 0.80 mm. There is a current of 2.5 mA in the 1.60- mm-diameter section. (a) What is the current in the 0.80-mm- diameter section? (b) What is the magnitude of E in the 1.60-mm-diameter section? (c) What is the magnitude of E in the 0.80-mm-diameter section? (d) What is the potential difference between the ends of the 3.00-m length of wire? Expert solution Step 1 Given We are given two lengths with two different radii for the same wire where 21 = 1.2m. 71 = 0.8 mm, L2 = 1.8m, r2 = 0.4 mm. Where the current through the first section is /1 = 2.5 mA. Solution (a) We want to determine the current flows in the second section /2. Both sections are connected in a series, where the resistance changes due to the change in radius. Also, the voltage changes, so the current will be the same for both sections and the current in the section with diameter 0.80 mm is 40.80 = 2.5 mA Step 2 (b) We want to determine the electric field E1. The greater the resistivity, the greater the electric field to cause a given current where the electric field is related to the resistivity by equation 25.5 in the form E = pJ = p= = 0 (1) Where p is the resistivity of the copper and equals 1.72 x 10-$ 0 . m (See Table 25.1 ) Now we can plug our values for p, I and 71 into equation (1) to get E1 E - pl - (1.72 x 10 8 42 . m) (0.0025A) (0.0008m)2 = 214 #V/m The electric field through the section with diameter 1.6 mm is 214 pV/m Step 3 (c) With the same steps in part (b) we could find the electric field for the section with diameter 0.80 mm. Let us plug our values for p. I and ry into equation (1) to get E2 E, = pl (1.72 x 10-8 0 . m) (0.0025A) ar3 (0.0004m) 85.5 AV/m The electric field through the section with diameter 0.80 mm is 85.5 pV/m. As shown by the results, as the radius is halved, the electric field increases four times. Step 4 (d) We want to determine the voltage across the entire wire. The voltage is across both sections, where current is the same but the resistance changes. So we could use Ohm's law in the next form to find the voltage V = I(R1 + R2) (Use Eq. 25.10) V = Im V = - Ip Li Le (2) Now we can plug our values for p, I, Ly, ry, Ly and re into equation (2) to get the voltage across the entire wire (0.0025mA) (1.72 x 10-$ 0 . m) 1.2 m 1.8 m (0.0008m)2 (0.0004m)2 = 0.180mV