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

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Current Electricity

The current density in a cylindrical wire of radius 4 mm is

4 \times 1 0^{6} A m^{- 2}

. The current through the outer portion of the wire between radial distance

\frac{R}{2}

and R is ______

π A

Answer: 48

Current density

J

relates to current by

I = J \cdot A

, where

A

is the cross-sectional area.

The area of the outer annular region between

r = \frac{R}{2}

and

r = R

is:

A = π R^{2} - π (\frac{R}{2})^{2} = π R^{2} (1 - \frac{1}{4}) = \frac{3 π R ^{2}}{4}

Given

R = 4 \times 1 0^{- 3}

m and

J = 4 \times 1 0^{6}

A/m²:

I = 4 \times 1 0^{6} \times \frac{3 π ( 4 \times 1 0 ^{- 3} ) ^{2}}{4} = 4 \times 1 0^{6} \times \frac{3 π \times 16 \times 1 0 ^{- 6}}{4} = 48 π A

Therefore, the answer is 48.

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Integer answer