A copper wire of length 1 m and radius 1 mm | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Current density $J$ is defined as $J = I / A$,where $I$ is the current and $A$ is the cross-sectional area.Since the wires are connected in series

Vedclass · Accepted Answer

$9 : 1$

Vedclass · Answer

(B) Current density $J$ is defined as $J = I / A$,where $I$ is the current and $A$ is the cross-sectional area.Since the wires are connected in series,the current $I$ flowing through both wires is the same.The cross-sectional area $A$ is given by $A = \pi r^2$.For the copper wire $(1)$: $J_1 = I / (\pi r_1^2)$.For the iron wire $(2)$: $J_2 = I / (\pi r_2^2)$.The ratio of current densities is $J_1 / J_2 = (I / \pi r_1^2) / (I / \pi r_2^2) = r_2^2 / r_1^2$.Given $r_1 = 1 \ mm$ and $r_2 = 3 \ mm$,we have:$J_1 / J_2 = (3)^2 / (1)^2 = 9 / 1$.Therefore,the ratio is $9 : 1$.

$A$ copper wire of length $1 \ m$ and radius $1 \ mm$ is connected in series with an iron wire of length $2 \ m$ and radius $3 \ mm$. If a current flows through both wires,the ratio of the current density in the copper wire to that in the iron wire will be:

Similar Questions

$A$ current $I$ flows through a uniform wire of diameter $d$ when the mean electron drift velocity is $V$. The same current will flow through a wire of diameter $d/2$ made of the same material if the mean drift velocity of the electron is:

When there is an electric current through a conducting wire along its length,then an electric field must exist

The current density in a cylindrical wire of radius $4 \; mm$ is $4 \times 10^{6} \; A \cdot m^{-2}$. The current through the outer portion of the wire between radial distance $\frac{R}{2}$ and $R$ is $\dots \; \pi \; A$.

$A$ current passes through a wire of nonuniform cross-section. Which of the following quantities are independent of the cross-section?

$A$ copper wire of length $10 \, m$ and radius $(10^{-2} / \sqrt{\pi}) \, m$ has an electrical resistance of $10 \, \Omega$. The current density in the wire for an electric field strength of $10 \, V/m$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module