Answer: E = 39.7 N/C
Explanation:
Using the equation
E= i/nAu
Where E is the magnitude of electric field in the wire, A is the cross sectional area of the wire that electron move through, n is the density of mobile electron, u is the mobility of the electrons due to the applied electric field, i is the number of electrons that enter the wire every second.
i is given 3× 10^18 e/s
A = 0.01mm2 = 1×10^-8 m2
And the values for n and u are given.
Substituting the values into the equation we have.
E = 3×10^18/(6.3 × 10^28 e/m3)( 1×10^-8 m2)( 1.2 × 10^-4m/s(N/C))
E= 39.7N/C
Answer:
This is a typical ohm' law question on mobile electrons in an electric field E within a tungsten filament.
Recall that the drifting velocity u, of mobile electrons in this field can be expressed as
u = μ x E ........................ Eqn 1
where;
μ = mobility of electron carriers = 1.2 x 10⁻⁴ (m/s)/(N/C)
Recall, the Electric current I, number of electrons per second, can also be expressed as
I = n x A x u ........................ Eqn 2
Where I = 3 x 10¹⁸ e/s,
A = cross-sectional area of the filament wire = 1mm² = 1 x 10⁻⁶m²
n = no of electrons per unit volume = 6.3 x 10²⁸ e/m³
Substituting 'u' in Eqn 1 into Eqn 2 gives,
E = [tex]\frac{I}{{n}{A}{μ}}[/tex]
So, E = [tex]\frac{3 x 10¹⁸}{{6.3 x 10²⁸}{1 x 10⁻⁸}{1.2 x 10⁻⁴}}[/tex]
E = 39.7 N/C
Hence, the magnitude of the electric field inside the tungsten filament in bulb B1 is 39.7 N/C.
