JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Current Electricity

The current density in a cylindrical wire of radius

r = 4.0

mm is

1.0 \times 1 0^{6}

A/m

^{2}

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

r /2

and r is

x π

A; where x is ______.

Answer: 12

Solution

The current through a differential ring element at radius

ρ

with thickness

d ρ

is:

d I = J \cdot d A = J \cdot 2 π ρ d ρ

Integrating from

r /2

r

I = \int_{r /2}^{r} J \cdot 2 π ρ d ρ = 2 π J [\frac{ρ ^{2}}{2}]_{r /2}^{r}

I = π J [r^{2} - \frac{r ^{2}}{4}] = π J \cdot \frac{3 r ^{2}}{4}

Substituting

J = 1.0 \times 1 0^{6}

A/m

^{2}

and

r = 4.0 \times 1 0^{- 3}

I = π \times 1 0^{6} \times \frac{3 \times ( 4 \times 1 0 ^{- 3} ) ^{2}}{4} = π \times 1 0^{6} \times 3 \times 4 \times 1 0^{- 6} = 12 π A

Therefore,

x = 12

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Integer answer