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 distances $\frac{R}{2}$ and R is _________πA.

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Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

J (current density) = 4 × 10⁶ Am^-2

Area between radial distance $\frac{R}{2} t o R$

A = $π [R^{2} - \frac{R^{2}}{4}] = \frac{3 R^{2}}{4} π$

I = AJ

= $\frac{3 R^{2}}{4} π \times 4 \times 1 0^{6}$

= 3R² × 10⁶ πAmp

= 3 (4 × 10^-3)² × 10⁶ πAmp

= 3 × 16 × 10^-6 × 10⁶π Amp

= 48πA

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