A rectangular copper strip 1.5cm wide and 0.10cn thick carries a current of 5.0A. Find the Hall voltage for a 1.2T magnetic fiel
1 answer:
Answer:
4.4345× 10^-7V
Explanation:
The computation of the half voltage for a 1.2T magnetic field applied is shown below
The volume of one mole of copper is
v = m ÷p
= 63.5 ÷ 8.92
= 7.12cm
Now the density of free electrons in copper is
n = Na ÷ V
= 6.02 × 10^23 ÷ 7.12
= 8.456× 10^28/m^3
Now the half voltage is
= IB ÷ nqt
= (5 × 1.20) ÷ (8.456× 10^28 × 1.6 × 10^-19 × 0.1× 10^-2)
= 4.4345× 10^-7V
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<span>C) either the pressure of the gas, the volume of the gas, or both, will increase.
In fact, the ideal gas law can be written as
</span>
<span>where
p is the gas pressure
V is its volume
n is the number of moles
R is the gas constant
T is the absolute temperature of the gas
We can see that if the temperature T increases, then the term on the right in the equation increases, therefore the term on the left should increase as well. In order for this to be possible, at least one between p and V should increase, or also both of them. Therefore, the correct answer is C.</span>
<span>C) either the pressure of the gas, the volume of the gas, or both, will increase.
In fact, the ideal gas law can be written as
</span>
<span>where
p is the gas pressure
V is its volume
n is the number of moles
R is the gas constant
T is the absolute temperature of the gas
We can see that if the temperature T increases, then the term on the right in the equation increases, therefore the term on the left should increase as well. In order for this to be possible, at least one between p and V should increase, or also both of them. Therefore, the correct answer is C.</span>