The current density in a cylindrical wire of radius r =4.0 mm is 1.0 × 106 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

Question

The current density in a cylindrical wire of radius r =4.0 mm is 1.0 × 106 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  (A)  (B)  (C)  (D)

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<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/2fa257c44c61cd89001a9e0a7f451325-.png" /> I=∫ JdA =∫ 106 × 2 π xdx .=106 × 2 π ⋅ x ( x 2/2)]( r /2) r =π × 106[ r 2-( r 2/4)]=12 π x =12

Q. 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 ______

$I = \int Jd A$
$= \int 1 0^{6} \times 2 π x d x$
$= 1 0^{6} \times 2 π \cdot x \frac{x ^{2}}{2}]_{\frac{r}{2}}^{r}$
$= π \times 1 0^{6} [r^{2} - \frac{r ^{2}}{4}] = 12 π$
$x = 12$

Q. The current density in a cylindrical wire of radius r=4.0mm is 1.0×106A/m2. The current through the outer portion of the wire between radial distances r/2 and r is xπA; where x is ______

I=∫JdA =∫106×2πxdx =106×2π⋅x2x2​]2r​r​ =π×106[r2−4r2​]=12π x=12

Q. 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 ______

$I = \int Jd A$
$= \int 1 0^{6} \times 2 π x d x$
$= 1 0^{6} \times 2 π \cdot x \frac{x ^{2}}{2}]_{\frac{r}{2}}^{r}$
$= π \times 1 0^{6} [r^{2} - \frac{r ^{2}}{4}] = 12 π$
$x = 12$