A cylindrical wire of radius 0.5 , mm and con | vedclass

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

Question

(A) Given:Conductivity $\sigma = 5 	imes 10^{7} \, S/m$Radius $r = 0.5 \, mm = 5 	imes 10^{-4} \, m$Electric field $E = 10 \, mV/m = 10 	imes 10^{-

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) Given:Conductivity $\sigma = 5 	imes 10^{7} \, S/m$Radius $r = 0.5 \, mm = 5 	imes 10^{-4} \, m$Electric field $E = 10 \, mV/m = 10 	imes 10^{-3} \, V/m = 10^{-2} \, V/m$Using Ohm's law in microscopic form,current density $J = \sigma E$:$J = (5 	imes 10^{7}) 	imes (10^{-2}) = 5 	imes 10^{5} \, A/m^{2}$Current $i = J 	imes A = J 	imes (\pi r^{2})$:$i = (5 	imes 10^{5}) 	imes \pi 	imes (5 	imes 10^{-4})^{2}$$i = 5 	imes 10^{5} 	imes \pi 	imes 25 	imes 10^{-8}$$i = 125 	imes 10^{-3} \, \pi \, A$$i = 125 \, \pi \, mA$Given that $i = x^{3} \, \pi \, mA$,we have:$x^{3} = 125$$x = \sqrt[3]{125} = 5$

$A$ cylindrical wire of radius $0.5 \, mm$ and conductivity $5 \times 10^{7} \, S/m$ is subjected to an electric field of $10 \, mV/m$. The expected value of current in the wire will be $x^{3} \, \pi \, mA$. The value of $x$ is ......

Similar Questions

The electric field $E$,current density $J$ and conductivity $\sigma$ of a conductor are related as

Assertion: Free electrons always keep on moving in a conductor,even then no magnetic force acts on them in a magnetic field unless a current is passed through it.
Reason: The average velocity of free electrons is zero.

Derive the relation between electric current and drift velocity.

Define mobility and write its $SI$ unit.

Current density in a cylindrical wire of radius $R$ varies with radial distance as $J(r) = \beta(r + r_0)^2$. The current through the shaded section of the wire shown in the figure is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module