Question

A copper wire of radius 0.6 mm is connected in series with another copper wire of radius 0.5 mm, and
they carry current of 2.4 A. The ratio of current density in the thinner wire to that in the thicker wire is,

Answer 1

Answer:

Radius of aluminium wire=1.25×10−3m

Cross-sectional area of aluminium=πr2

=4.9×10−6m

∴J=4.9×10−610=2.04×106Am−2

hope this will help you

Answer 2

Answer:

The ratio of current density in the thinner wire to that in the thicker wire is equal to 1.44.

Explanation:

Current density in the wire is the current flowing through the wire per unit area of cross section at a point.

Current density = Current /area of cross section

$J=\frac{I}{A}$

We have the radius of first copper wire (thicker) $r_{1}= 0.6mm = 0.6*10^{-3}m$

The radius of the second copper wire (thinner), $r_{2}= 0.5mm=0.5*10^{-3}m$

The current flowing, $I=2.4A$

Cross section area of thicker copper wire $A_{1}=\pi r_{1}^2=(3.14)(0.6*10^{-3})^2$

$A_{1}=1.1304*10^{-6}m^2$

Cross section area of thinner copper wire $A_{2}=\pi r_{2}^2=(3.14)(0.5*10^{-3})^2$

$A_{1}=0.785*10^{-6}m^2$

The ratio of current density in the thinner wire (J₂) to that in the thicker wire (J₁):-

$\frac{\big J_{2}}{\big J_{1}} =\frac{\big I/\big A_{2}}{\big I/\big A_{1}}$

$\frac{\big J_{2}}{\big J_{1}} =\frac{\big A_{2}}{\big A_{1}}$

$\frac{\big J_{2}}{\big J_{1}} =\frac{1.1304* 10^{-6}}{0.785*10^_{-6}}$

$\frac{J_{2}}{J_{1}} =1.44$

Therefore, the ration of current density in the thinner wire to that in the thicker wire  will be equal to 1.44.