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3 years ago

8

The Hall effect can be used to determine the density of mobile electrons in a conductor. A thin strip of the material being inve

stigated is immersed in a magnetic field and oriented so that its surface is perpendicular to the field. In a particular measurement, the magnetic field strength was 0.685 T, the strip was 0.107 mm thick, the current along the strip was 2.25 A, and the Hall voltage between the strip's edges was 2.59 mV.Find the density nof mobile electrons in the material. The elementary charge is 1.602×10−19 C.

1 answer:

solmaris [256]3 years ago

5 0

Answer:

the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³

Explanation:

Given the data in the question;

we make use of the following expression;

hall Voltage VH = IB / ned

where I = 2.25 A

B = 0.685 T

d = 0.107 mm = 0.107 × 10⁻³ m

e = 1.602×10⁻¹⁹ C

VH = 2.59 mV = 2.59 × 10⁻³ volt

n is the electron density

so from the form; VH = IB / ned

VHned = IB

n = IB / VHed

so we substitute

n = (2.25 × 0.685) / ( 2.59 × 10⁻³ × 1.602×10⁻¹⁹ × 0.107 × 10⁻³ )

n = 1.54125 / 4.4396226 × 10⁻²⁶

n = 3.4716 × 10²⁵ m⁻³

Therefore, the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³

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A charge of −20 µC is distributed uniformly over the surface of a spherical conductor of radius 11.0 cm. Determine the electric

Alex73 [517]

Answer:

(a) $-6.76\times 10^{12}\ N/C$

(b) $-1.352\times 10^{13}\ N/C$

(c) $-7.2\times 10^{11}\ N/C$

Explanation:

(a)

Given:

Charge on sphere (Q) = $-20\ \mu C=-20\times 10^{-6}\ C$

Radius of sphere (R) = 11.0 cm = 0.110 m

Distance from the center (r) = 5 cm = 0.05 m

Coulomb's constant (k) = $9\times 10^{9}\ Nm^2/C^2$

Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r ≤ R' from the center of sphere is given as:

$E=(\frac{kQ}{R^3})r$

Plug in the given values and solve for 'E'. This gives,

$E_{in}=(\frac{9\times 10^{9}\times -20}{(0.110)^3})\times 0.05\\\\E_{in}=-1.352\times 10^{14}\times 0.05\\\\E_{in}=-6.76\times 10^{12}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)$

(b)

Given:

Charge on sphere (Q) = $-20\ \mu C=-20\times 10^{-6}\ C$

Radius of sphere (R) = 11.0 cm = 0.110 m

Distance from the center (r) = 10 cm = 0.10 m

Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r ≤ R' from the center of sphere is given as:

$E=(\frac{kQ}{R^3})r$

Plug in the given values and solve for 'E'. This gives,

$E_{in}=(\frac{9\times 10^{9}\times -20}{(0.110)^3})\times 0.10\\\\E_{in}=-1.352\times 10^{14}\times 0.10\\\\E_{in}=-1.352\times 10^{13}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)$

(c)

Given:

Charge on sphere (Q) = $-20\ \mu C=-20\times 10^{-6}\ C$

Radius of sphere (R) = 11.0 cm = 0.110 m

Distance from the center (r) = 50 cm = 0.50 m

Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r > R' from the center of sphere is given as:

$E=\dfrac{kQ}{r^2}$

Plug in the given values and solve for 'E'. This gives,

$E_{out}=(\frac{9\times 10^{9}\times -20}{(0.50)^2})\\\\E_{out}=-7.2\times 10^{11}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)$

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4 years ago

Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV

Korolek [52]

Complete Question

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).

Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV reciever consisting of a circular dish of radius R which focuses the electromagnetic energy incident from the satellite onto a receiver which has a surface area of 5 cm2.

How large does the radius R of the dish have to be to achieve an electric field vector amplitude of 0.1 mV/m at the receiver?

For simplicity, assume that your house is located directly beneath the satellite (i.e. the situation you calculated in the first part), that the dish reflects all of the incident signal onto the receiver, and that there are no losses associated with the reception process. The dish has a curvature, but the radius R refers to the projection of the dish into the plane perpendicular to the direction of the incoming signal.

Give your answer in centimeters, to two significant figures.

Answer:

The radius of the dish is $R = 18cm$

Explanation:

From the question we are told that

The radius of the orbit is = $R = 35,000km = 35,000 *10^3 m$

The power output of the power is $P = 1 kW = 1000W$

The electric vector amplitude is given as $E = 0.1 mV/m = 0.1 *10^{-3}V/m$

The area of thereciever is $A_R = 5cm^2$

Generally the intensity of the dish is mathematically represented as

$I = \frac{P}{A}$

Where A is the area orbit which is a sphere so this is obtained as

$A = 4 \pi r^2$

$= (4 * 3.142 * (35,000 *10^3)^2)$

$=1.5395*10^{16} m^2$

Then substituting into the equation for intensity

$I_s = \frac{1000}{1.5395*10^{16}}$

$= 6.5*10^ {-14}W/m2$

Now the intensity received by the dish can be mathematically evaluated as

$I_d = \frac{1}{2} * c \epsilon_o E_D ^2$

Where c is thesped of light with a constant value $c = 3.0*10^8 m/s$

$\epsilon_o$ is the permitivity of free space with a value $8.85*10^{-12} N/m$

$E_D$ is the electric filed on the dish

So since we are to assume to loss then the intensity of the satellite is equal to the intensity incident on the receiver dish

Now making the eletric field intensity the subject of the formula

$E_D = \sqrt{\frac{2 * I_d}{c * \epsilon_o} }$

substituting values

$E_D = \sqrt{\frac{2 * 6.5*10^{-14}}{3.0*10^{8} * 8.85*10^{-12}} }$

$= 7*10^{-6} V/m$

The incident power on the dish is what is been reflected to the receiver

$P_D = P_R$

Where $P_D$ is the power incident on the dish which is mathematically represented as

$P_D = I_d A_d$

$= \frac{1}{2} c \epsilon_o E_D^2 (\pi R^2)$

And $P_R$ is the power incident on the dish which is mathematically represented as

$P_R = I_R A_R$

$= \frac{1}{2} c \epsilon_o E_R^2 A_R$

Now equating the two

$\frac{1}{2} c \epsilon_o E_D^2 (\pi R^2) = \frac{1}{2} c \epsilon_o E_R^2 A_R$

Making R the subject we have

$R = \sqrt{\frac{E_R^2 A_R}{\pi E_D^2} }$

Substituting values

$R = \sqrt{\frac{(0.1 *10^{-3})^2 * 5}{\pi (7*10^{-6})^ 2} }$

$R = 18cm$

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