(B) The current $I$ is given by the dot product of current density $\vec{J}$ and area vector $\vec{A}$: $I = \vec{J} \cdot \vec{A} = J A \cos(\theta)$

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

(B) The current $I$ is given by the dot product of current density $\vec{J}$ and area vector $\vec{A}$: $I = \vec{J} \cdot \vec{A} = J A \cos(\theta)$.Given $I = 5\; A$,$A = 0.04\; m^2$,and $\theta = 60^{\circ}$.Substituting the values: $5 = J \times 0.04 \times \cos(60^{\circ})$.Since $\cos(60^{\circ}) = 0.5$,we have $5 = J \times 0.04 \times 0.5 = J \times 0.02$.Thus,$J = \frac{5}{0.02} = 250\; A/m^2$.Using Ohm's law in vector form,$\vec{E} = \rho \vec{J}$,the magnitude is $E = \rho J$.$E = (44 \times 10^{-8}) \times 250 = 11000 \times 10^{-8} = 11 \times 10^{-5}\; V/m$.

Class 12 Physics Current Electricity Current Density, Drift Velocity and Mobility

Difficult

$A$ current of $5\; A$ is passing through a non-linear magnesium wire of cross-section $0.04\; m^2$. At every point,the direction of current density is at an angle of $60^{\circ}$ with the unit vector of the area of cross-section. The magnitude of the electric field at every point of the conductor is .... $V/m$ (Resistivity of magnesium is $\rho = 44 \times 10^{-8}\, \Omega m$).

JEE MAIN 2021 All JEE MAIN

A
$11 \times 10^{-3}$
B
$11 \times 10^{-5}$
C
$11 \times 10^{-7}$
D
$11 \times 10^{-2}$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Physics Questions

Chapter

Current Electricity

Subtopic

Current Density, Drift Velocity and Mobility

Previous Year

JEE MAIN 2021 Paper All JEE MAIN years →

$A$ conductor of length $100 \, cm$ and area of cross-section $1 \, mm^2$ carries a current of $5 \, A$. If the resistivity of the material of the conductor is $3.0 \times 10^{-8} \, \Omega \cdot m$, then the electric field across the conductor is (in $ \, V/m$)

MediumTS EAMCET 2020

View Solution

$A$ cylindrical conductor of length $2 \ m$ and area of cross-section $0.2 \ mm^{2}$ carries an electric current of $1.6 \ A$ when its ends are connected to a $2 \ V$ battery. Mobility of electrons in the conductor is $\alpha \times 10^{-3} \ m^{2}/V \cdot s$. The value of $\alpha$ is: (electron concentration $n = 5 \times 10^{28} \ m^{-3}$ and electron charge $e = 1.6 \times 10^{-19} \ C$)

MediumJEE MAIN 2026

View Solution

The drift velocity of electrons in a silver wire with a cross-sectional area of $3.14 \times 10^{-6} \, m^2$ carrying a current of $20 \, A$ is. Given the atomic weight of $Ag = 108$ and the density of silver $= 10.5 \times 10^3 \, kg/m^3$,the drift velocity is $.......... \times 10^{-4} \, m/s$.

Easy

View Solution

When $5\ V$ potential difference is applied across a wire of length $0.1\ m$,the drift speed of electrons is $2.5 \times 10^{-4} \ m/s$. If the electron density in the wire is $8 \times 10^{28} \ m^{-3}$,the resistivity of the material is close to:

MediumJEE MAIN 2015

View Solution

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.

EasyAIIMS 2016

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Similar Questions

$A$ conductor of length $100 \, cm$ and area of cross-section $1 \, mm^2$ carries a current of $5 \, A$. If the resistivity of the material of the conductor is $3.0 \times 10^{-8} \, \Omega \cdot m$, then the electric field across the conductor is (in $ \, V/m$)

The drift velocity of electrons in a silver wire with a cross-sectional area of $3.14 \times 10^{-6} \, m^2$ carrying a current of $20 \, A$ is. Given the atomic weight of $Ag = 108$ and the density of silver $= 10.5 \times 10^3 \, kg/m^3$,the drift velocity is $.......... \times 10^{-4} \, m/s$.

When $5\ V$ potential difference is applied across a wire of length $0.1\ m$,the drift speed of electrons is $2.5 \times 10^{-4} \ m/s$. If the electron density in the wire is $8 \times 10^{28} \ m^{-3}$,the resistivity of the material is close to:

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.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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.