Respuesta :
Answer:
The current in the filament is 0.81 A
Explanation:
Given :
Diameter of wire [tex]d_{1} = 0.12 \times 10^{-3}[/tex] m
Diameter of filament [tex]d_{2} = 1.5 \times 10^{-3}[/tex] m
Current density [tex]J = 4.6 \times 10^{5} \frac{A}{m^{2} }[/tex]
From the formula of current density,
  [tex]J = \frac{I}{A}[/tex]
Where [tex]A =[/tex] area = [tex]\frac{\pi d^{2} }{4}[/tex]
Current in the filament is given by,
  [tex]I = JA[/tex]
Area of filament is given by [tex]A = \pi \frac{(1.5 \times 10^{-3} )^{2} }{4}[/tex]
 [tex]A = 1.76 \times 10^{-6}[/tex]
Put value of A in above equation,
 [tex]I = 4.6 \times 10^{5} \times 1.76 \times 10^{-6}[/tex]
 [tex]I = 0.81[/tex] A
Therefore, current in the filament is 0.81 A
