Ask Your Teacher A block in the shape of a rectangular solid has a cross-sectional area of 3.00 cm2 across its width, a front-to-rear length

Question

Ask Your Teacher A block in the shape of a rectangular solid has a cross-sectional area of 3.00 cm2 across its width, a front-to-rear length of 15.1 cm, and a resistance of 945 Ω. The block’s material contains 5.33 ✕ 1022 conduction electrons/m3. A potential difference of 35.4 V is maintained between its front and rear faces. (a) What is the current in the block?

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Kiệt Gia 1 month 2021-08-12T13:52:46+00:00 2 Answers 1 views 0

Answers ( )

    0
    2021-08-12T13:54:15+00:00

    Answer:

    Explanation:

    Given:

    Cross-sectional area, A = 3.00 cm^2

    Converting from cm^2 to m^2,

    3 cm^2 × (1 m)^2/(100 cm)^2

    = 3 × 10^-4 m^2

    Length, L = 15.1 cm

    Resistance, R = 945 Ω

    N = 5.33 ✕ 10^22 conduction electrons/m^3

    Potential difference, V = 35.4 V

    A.

    Using ohm’s law,

    Potential difference, V = current, I × resistance, R

    I = 35.4/945

    = 0.0375 A.

    B.

    Current density, J = I/A

    = 0.0375/3 × 10^-4

    = 124.87 A/m^2

    0
    2021-08-12T13:54:39+00:00

    Answer:

    a. 37.5 mA b. 124.9 A/m² c. 1.46 cm/s d. 234.44 V/m

    Explanation:

    Here is the complete question

    A block in the shape of a rectangular solid has a cross-sectional area of 3.00 cm2 across its width, a front-to-rear length of 15.1 cm, and a resistance of 945 Ω. The block’s material contains 5.33 × 1022 conduction electrons/m3. A potential difference of 35.4 V is maintained between its front and rear faces. (a) What is the current in the block? (b) If the current density is uniform, what is its magnitude? What are (c) the drift velocity of the conduction electrons and (d) the magnitude of the electric field in the block?

    Solution

    a. The current in the block, I

    I = V/R where V = potential difference between its front and rear faces = 35.4 V, R = resistance = 945 Ω

    I = V/R = 35.4 V/945 Ω = 0.0375 A = 37.5 mA

    b. The current density J

    J = I/A  where A = cross-sectional area of block = 3.00 cm² = 3 × 10⁻⁴ m²

    J = I/A = 0.0375 A /3 × 10⁻⁴ m² = 124.9 A/m²

    c. The drift velocity of the electrons

    From J = nev, v = J/ne where n = electron density = 5.33 × 10²² electrons/m³

    e = electron charge = 1.602 × 10⁻¹⁹C

    v = J/ne = 124.9 A/m²/(5.33 × 10²² electrons/m³ × 1.602 × 10⁻¹⁹C)

      = 0.0146 m/s = 1.46 cm/s

    d. The magnitude of the electric field, E

    E = V/L where V = potential difference applied between its front and rear ends = 35.4 V and L = front-to-rear length of block = 15.1 cm = 0.151 m

    E = V/L = 35.4 V/0.151 m = 234.44 V/m

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