Question

The wires leading to and from a 0.12-mm-diameter lightbulb filament are 1.5 mm in diameter. The wire to the filament carries a current with a current density of 4.6×105 A/m2 . What is the current in the filament? Express your answer to two significant figures and include the appropriate units.

Answer 1

Answer:

The current in the filament is 0.81 A

Explanation:

Given :

Diameter of wire $d_{1} = 0.12 \times 10^{-3}$ m

Diameter of filament $d_{2} = 1.5 \times 10^{-3}$ m

Current density $J = 4.6 \times 10^{5} \frac{A}{m^{2} }$

From the formula of current density,

   $J = \frac{I}{A}$

Where $A =$ area = $\frac{\pi d^{2} }{4}$

Current in the filament is given by,

   $I = JA$

Area of filament is given by $A = \pi \frac{(1.5 \times 10^{-3} )^{2} }{4}$

  $A = 1.76 \times 10^{-6}$

Put value of A in above equation,

  $I = 4.6 \times 10^{5} \times 1.76 \times 10^{-6}$

  $I = 0.81$ A

Therefore, current in the filament is 0.81 A