The current density in a cylindrical wire of radius 4 ; mm is 4 times 10^{6} ; Am ^{-2}. The current through the outer portion

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

b $J=\frac{I}{A}$ $I=J A$ $=4 \times 10^{6} \times\left[\pi R^{2}-\pi\left(\frac{R}{2}\right)^{2}\right]$ $=4 \times 10^{6} \times \pi R^{2} \times \frac{3}{4}$ $=4 \times 10^{6} \times \pi \times\left(4 \times 10^{-3}\right)^{2} \times \frac{3}{4}=48 \; \pi A .$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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

A$-48$
B$48$
C$-58$
D$58$

JEE MAIN 2022, Medium

STD 11 Science
NEET
STD 12 - 3. current electricity
Physics

Download our app
and get started for free

Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*

Download App

Download our appand get started for free

Similar Questions

Download our app
and get started for free