Which would likely demagnetize a magnetized iron nail?

A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the magnitude of the dri

nordsb [41]

Complete Question

A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm2.) mm/s

Answer:

The drift velocity is $v = 0.0002808 \ m/s$

Explanation:

From the question we are told that

The current on the copper is $I = 20 \ A$

The cross-sectional area is $A = 5.261 \ mm^2 = 5.261 *10^{-6} \ m^2$

The number of copper atom in the wire is mathematically evaluated

$n = \frac{\rho * N_a}Z}$

Where $\rho$ is the density of copper with a value $\rho = 8.93 \ g/m^3$

$N_a$ is the Avogadro's number with a value $N_a = 6.02 *10^{23}\ atom/mol$

Z is the molar mass of copper with a value $Z = 63.55 \ g/mol$

So

$n = \frac{8.93 * 6.02 *10^{23}}{63.55}$

$n = 8.46 * 10^{28} \ atoms /m^3$

Given the 1 atom is equivalent to 1 free electron then the number of free electron is

$N = 8.46 * 10^{28} \ electrons$

The current through the wire is mathematically represented as

$I = N * e * v * A$

substituting values

$20 = 8.46 *10^{28} * (1.60*10^{-19}) * v * 5.261 *10^{-6}$

=> $v = 0.0002808 \ m/s$

8 0

3 years ago

A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur

muminat

Answer:

a) E = 0

b) $E = \dfrac{k_e \cdot q}{ r^2 }$

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

From which we have;

$E \cdot A = \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}} = \dfrac{0}{\varepsilon _{0}} = 0$

E = 0/A = 0

E = 0

b) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

$E \cdot A = \dfrac{+q }{\varepsilon _{0}}$

$E = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}$

By Gauss theorem, we have;

$E\oint dS = \dfrac{q}{\varepsilon _{0}}$

Therefore, we get;

$E \cdot (4 \cdot \pi \cdot r^2) = \dfrac{q}{\varepsilon _{0}}$

The electrical field outside the spherical shell

$E = \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }= \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }$

$k_e= \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }$

Therefore, we have;

$E = \dfrac{k_e \cdot q}{ r^2 }$

5 0

3 years ago

What is Vatiable Velocity?

Allushta [10]

Answer:

Explanation: When a body covers unequal distances in equal intervals of time in a specified direction, the body is said to be moving with a variable velocity. Example: A rotating fan at a constant speed has variable velocity, because of continuous change in direction.

7 0

3 years ago

What is an ideal diode? a. The ideal diode acts as an open circuit for forward currents and as a short circuit with reverse volt

Serhud [2]

Answer:

d. The ideal diode acts as a short circuit for forward currents and as an open circuit with reverse voltage applied.

Explanation:

Ideal diode acts like an ideal conductor. In case of forward voltage it acts like an ideal conductor. However when it is reverse biased then it behaves like an ideal insulator. You can understand it bu considering a switch. When the voltage is forward then ideal diode acts like a closed switch. When the voltage is reverse biased then ideal diode behaves like an open switch.

That is why we can say that the ideal diode acts as a short circuit (higher conduction) for forward currents and as an open circuit ( zero conduction) with reverse voltage applied.

8 0

3 years ago

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.60×106 N, one at an angle 13.0 west of north, an

Juliette [100K]

Answer: $W=1.93\times 10^9 J$