The radius of the thinner wire is 0.22 mm and the radius of the thick wire is 0.55 mm. There are 4.0 x 1028 mobile electrons per cubic meter

Question

The radius of the thinner wire is 0.22 mm and the radius of the thick wire is 0.55 mm. There are 4.0 x 1028 mobile electrons per cubic meter of this material and the electron mobility is 6.0 x 10-4 (m/s)/(N/C). If 6.0 x 1018 electrons pass location D each second, what is the magnitude of the electric field at location B

in progress 0
Delwyn 5 months 2021-08-06T13:08:38+00:00 1 Answers 4 views 0

Answers ( )

    0
    2021-08-06T13:09:41+00:00

    Answer:

    0.2631 N/C

    Explanation:

    Given that:

    The radius of the wire r = 0.22 mm = 0.22 × 10⁻³ m

    The radius of the thick wire  r’ = 0.55 mm = 0.55 × 10⁻³ m

    The numbers of electrons passing through B, N = 6.0  × 10¹⁸ electrons

    Electron mobility  μ =  6.0 x 10-4 (m/s)/(N/C)

    = 0.0006

    The number of electron flow per second is calculated as follows:

    I = \frac{q}{t}

    I = \frac{Ne}{t}

    I = \frac{6*10^{18}(1.6*10^{-18})C}{1 \ s}

    I = 0.96 \ A

    The magnitude of the electric field is:

    E = \frac{I}{ \mu n eA}

    E = \frac{I}{ \mu n e(\pi r^2)}

    E = \frac{0.96}{(0.0006 m/s N/C ) (4*10^{28})(1.6*10^{-19}C)(0.55*10^{-3}m)^2}

    E = 0.2631 N/C

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )