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STatiana [176]
3 years ago
15

5. The wire in consists of two segments of different diameters but made from the same metal. The current in segment 1 is I1. a.

(1 pt) Compare the currents in the two segments. That is, is I2 greater than, less than, or equal to I1? Explain. b. (1 pt) Compare the current densities J1 and J2 in the two segments. c. (1 pt) Compare the electric field strengths E1 and E2 in the two segments. d. (1 pt) Compare the drift speeds vd1 and vd2 in the two segments.
Physics
1 answer:
Volgvan3 years ago
7 0

Answer:

hello your question is incomplete attached below is the complete question

answer :

a) I1 = I2

b) J1 > J2

c) E 1 > E2

d) ( vd1 ) > ( vd2 )

Explanation:

a) The currents in the two segments are the same  i.e. I1 = I2  and this is because the segments are connected in series

b) Comparing the current densities J1 and J2 in the two segments

note : current density ∝ 1 / area

The area of the second segment is > the area of first segment  therefore

J1 > J2

J1 ( current density of first segment )

J2 ( current density of second segment )

c) Comparing the electric field strengths E1 and E2

 note : electric field strength ∝ current density

since current density of first segment is > current density of second segment  and conductivity of the materials are the same hence

E 1 > E2

d) Comparing the drift speeds Vd1 and Vd2

( vd1 ) > ( vd2 )

this because  ; vd ∝ current density

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A particle moves along line segments from the origin to the points (1, 0, 0), (1, 5, 1), (0, 5, 1), and back to the origin under
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Answer:

0 J

Explanation:

Since work done W = ∫F.dr and F(x, y, z)= z²i + 4xyj + 5y²k and dr = dxi + dyj + dzk

F.dr = (z²i + 4xyj + 5y²k).(dxi + dyj + dzk) = z²dx + 4xydy + 5y²dz

W = ∫F.dr = ∫z²dx + 4xydy + 5y²dz = z²x + 2xy² + 5y²z

We now evaluate the work done for the different regions

W₁ = work done from (0,0,0) to (1,0,0)

W₁ = {z²x + 2xy² + 5y²z}₀₀₀¹⁰⁰ = 0²(1) + 2(1)(0)² + 5(0)²(0) - [(0)²(0) + 2(0)(0)² + 5(0)²(0)] = 0 - 0 = 0 J

W₂ = work done from (1,0,0) to (1,5,1)

W₂ = {z²x + 2xy² + 5y²z}₁₀₀¹⁵¹ =   (1)²(1) + 2(1)(5)² + 5(5)²(1) - [0²(1) + 2(1)(0)² + 5(0)²(0)] =  1 + 50 + 125 - 0 = 176 J

W₃ = work done from (1,5,1) to (0,5,1)

W₃ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁵¹ =   1²(0) + 2(0)(5)² + 5(5)²(1) - [(1)²(1) + 2(1)(5)² + 5(5)²(1)]  = 125 - (1 + 50 + 125) = 125 - 176 = -51 J

W₄ = work done from (0,5,1) to (0,0,0)

W₄ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁰⁰ =   (0)²(0) + 2(0)(0)² + 5(0)²(0) - [1²(0) + 2(0)(5)² + 5(5)²(1)] = 0 - 125 = -125 J

The total work done W is thus

W = W₁ + W₂ + W₃ + W₄

W = 0 J + 176 J - 51 J - 125 J

W = 176 J - 176 J

W = 0 J

The total work done equals 0 J

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