To solve this problem it is necessary to use the concepts related to the Hall Effect and Drift velocity, that is, at the speed that an electron reaches due to a magnetic field.

The drift velocity is given by the equation:

$V_d = \frac{I}{nAq}$

Where

I = current

n = Number of free electrons

A = Cross-Section Area

q = charge of proton

Our values are given by,

$I = 25 A$

$A= 1.2*20 *10^{-6} m^2$

$q= 1.6*10^{-19}C$

$N = 8.47*10^{19} mm^{-3}$

$V_d =\frac{25}{(1.2*20 *10^{-6})(1.6*10^{-19})(8.47*10^{19} )}$

$V_d = 7.68*10^{-5}m/s$

The hall voltage is given by

$V=\frac{IB}{ned}$

Where

B= Magnetic field

n = number of free electrons

d = distance

e = charge of electron

Then using the formula and replacing,

$V=\frac{(2.5)(25)}{(8.47*10^{28})(1.6*10^{-19})(1.2*10^{-3})}$

$V = 3.84*10^{-6}V$

5 0

3 years ago

Which factor causes global wind patterns?

Kay [80]

Answer:

B . unequal heating of the Earth's surface by the sun

4 0

2 years ago

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A gas in a cylinder expands from a volume of 0.110 m³ to 0.320 m³. heat flows into the gas just rapidly enough to keep the press

Elden [556K]

80000 Joule is the change in the internal energy of the gas.

<h3>In Thermodynamics, work done by the gas during expansion at constant pressure:</h3>

ΔW = -pdV

ΔW = -pd (V₂ -V₁)

ΔW = - 1.65×10⁵ pa (0.320m³ - 0.110m³)

= - 0.35×10⁵ pa.m³

= - 35000 (N/m³)(m³)

= -35000 Nm

ΔW = -35000 Joule

Therefore, work done by the system = -35000 Joule

<h3>Change in the internal energy of the gas,</h3>

ΔV = ΔQ + ΔW

Given:

ΔQ = 1.15×10⁵ Joule

ΔW = -35000 Joule

ΔU = 1.15×10⁵ Joule - 35000 Joule

= 80000 Joule.

Therefore, the change in the internal energy of the gas= 80000 Joule.

Learn more about thermodynamics here:

brainly.com/question/14265296

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3 0

2 years ago