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Eduardwww [97]
3 years ago
13

A current of I = 3.8 A is passing through a conductor with cross sectional area A = 2.5 x 10^-4 m^2. The charge carriers in the

conductor, electrons, have a number density n = 1.3 x 10^27 m^-3.
(a) Express the drift velocity of electrons through 1.A. n, and the fundamental charge e.
(b) Calculate the numerical value of v, in m/s.
Physics
1 answer:
bulgar [2K]3 years ago
8 0

Answer:

Explanation:

current, i = 3.8 A

Area of crossection, A = 2.5 x 10^-4 m²

number density, n = 1.3 x 10^27 m^-3

(a) the relation between the current and the drift velocity is given by

i = n e A v

where, e is the electronic charge, A be the area of crossection area of the wire and v be the drift velocity

v = i / neA

(b)

3.8 = 1.3 x 10^27 x 1.6 x 10^-19 x 2.5 x 10^-4 x v

3.8 = 52000 v

v = 7.3 x 10^-5 m/s

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A short current element dl⃗ = (0.500 mm)j^ carries a current of 5.40 A in the same direction as dl⃗ . Point P is located at r⃗ =
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Answer:

The magnetic field along x axis is

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The magnetic field along z axis is

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Using Biot-savart law

B=\dfrac{\mu_{0}}{4\pi}\timesI\times\dfrac{\vec{dl}\times\vec{r}}{|\vec{r}|^3}

Put the value into the formula

B=10^{-7}\times5.40\times\dfrac{(0.5\times10^{-3})\times(-0.730)i+(0.390)k}{(0.827)^3}

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\vec{dl}\times\vec{r}=(0.5\times10^{-3})\times(-0.730)i+(0.390)k

\vec{dl}\times\vec{r}=i(0.350\times0.5\times10^{-3}-0)+k(0+0.730\times0.5\times10^{-3})

\vec{dl}\times\vec{r}=0.000175i+0.000365k

Put the value into the formula of magnetic field

B=10^{-7}\times5.40\times\dfrac{(0.000175i+0.000365k)}{(0.827)^3}

B=1.670\times10^{-10}i+3.484\times10^{-10}k

Hence, The magnetic field along x axis is

B_{x}=1.670\times10^{-10}\ T

The magnetic field along y axis is zero.

The magnetic field along z axis is

B_{z}=3.484\times10^{-10}\ T

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