The current density in a cylindrical wire of radius 4 mm is 4 × 106 Am -2. The current through the outer portion of the wire between radial distance (R/2) and R is π A .

Question

The current density in a cylindrical wire of radius 4 mm is 4 × 106 Am -2. The current through the outer portion of the wire between radial distance (R/2) and R is π A . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

J=(I/A) I=J A =4 × 106 ×[π R2-π((R/2))2] =4 × 106 × π R2 × (3/4) =4 × 106 × π ×(4 × 10-3)2 × (3/4)=48 π A .

Q. 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 _______ $π A .$

$J = \frac{I}{A}$
$I = J A$
$= 4 \times 1 0^{6} \times [π R^{2} - π (\frac{R}{2})^{2}]$
$= 4 \times 1 0^{6} \times π R^{2} \times \frac{3}{4}$
$= 4 \times 1 0^{6} \times π \times (4 \times 1 0^{- 3})^{2} \times \frac{3}{4} = 48 π A .$

Q. The current density in a cylindrical wire of radius 4mm is 4×106Am−2. The current through the outer portion of the wire between radial distance 2R​ and R is _______πA.

J=AI​ I=JA =4×106×[πR2−π(2R​)2] =4×106×πR2×43​ =4×106×π×(4×10−3)2×43​=48πA.

Q. 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 _______ $π A .$

$J = \frac{I}{A}$
$I = J A$
$= 4 \times 1 0^{6} \times [π R^{2} - π (\frac{R}{2})^{2}]$
$= 4 \times 1 0^{6} \times π R^{2} \times \frac{3}{4}$
$= 4 \times 1 0^{6} \times π \times (4 \times 1 0^{- 3})^{2} \times \frac{3}{4} = 48 π A .$