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 R 2 and R is _________πA.

J (current density) = 4 * 106 Am-2 Area between radial distance R 2 t o R A = π [ R 2 − R 2 4 ] = 3 R 2 4 π I = AJ = 3 R 2 4 π * 4 * 1 0 6 = 3R2 * 106 πAmp = 3 (4 * 10-3)2 * 106 πAmp = 3 * 16 * 10-6 * 106π Amp = 48πA

**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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Vishal Baghel | Contributor-Level 10

6 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

= $\frac{3 R^{2}}{4} π * 4 * 1 0^{6}$

= 3R² * 10⁶ πAmp

= 3 (4 * 10^-3)² * 10⁶ πAmp

= 3 * 16 * 10^-6 * 10⁶π Amp

= 48πA

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