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STD 12 - 3. current electricity — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3. current electricity4 Marks
MCQ
Current density in a cylindrical wire of radius $R$ is given as $J =$ $\left\{ {\begin{array}{*{20}{c}}
  {{J_0}\left( {\frac{x}{R} - 1} \right)\,\,for\,\,0 \leqslant x < \frac{R}{2}} \\ 
  {{J_0}\frac{x}{R}\,\,\,\,for\,\,\,\frac{R}{2} \leqslant x \leqslant R} 
\end{array}} \right.$The current flowing in the wire is:
  • A
    $\frac{7}{{24}}\pi {J_0}{R^2}$
  • B
    $\frac{1}{{6}}\pi {J_0}{R^2}$
  • C
    $\frac{7}{{14}}\pi {J_0}{R^2}$
  • ✓
    $\frac{5}{{12}}\pi {J_0}{R^2}$
✓

Answer

Correct option: D.
$\frac{5}{{12}}\pi {J_0}{R^2}$
d
$J=\left\{\begin{array}{ll}J_{0}\left(\frac{x}{R}-1\right) & 0 \leq x \leq \frac{R}{2} \\ J_{0} \frac{x}{R} & \frac{R}{2} \leq x \leq R\end{array}\right.$

we know that

$J=\frac{\text {current}}{\text {area}}$

$\Rightarrow$ current $=J$ Area

current $=J_{1} A_{1}+J_{2} A_{2}$

$=J_{0}\left(\frac{x}{R}-1\right) \times(2 \pi x d x)+J_{0} \frac{x}{R}(2 \pi x d x)$

$=\int_{0}^{R / 2} J_{0}\left(\frac{x}{R}-1\right) \times(2 \pi x d x)+\int_{R / 2}^{R} J_{0} \times 2 \pi \frac{x^{2}}{R} d x$

$=J_{0} \times 2 \pi \int_{0}^{R / 2}\left(\frac{x^{2}}{R}-x\right) d x+\frac{J_{0} \times 2 \pi}{R} \int_{R / 2}^{R} x^{2} d x$

$\Rightarrow i=J_{0} \times 2 \pi\left[\frac{x^{3}}{3 R}-\frac{x^{2}}{2}\right]_{0}^{R / 2}+\frac{J_{0} \times 2 \pi}{R}\left[\frac{x^{3}}{3}\right]_{R / 2}^{R}$

$\Rightarrow i=J_{0} \times 2 \pi\left[\frac{R^{2}}{24}-\frac{R^{2}}{8}\right]+\frac{J_{0} \times 2 \pi}{R}\left[\frac{R^{3}}{3}-\frac{R^{3}}{24}\right]$

$=J_{0} \times 2 \pi\left(\frac{-2 R^{2}}{24}\right)+\frac{J_{0} \times 2 \pi}{R}\left[\frac{7 R^{3}}{24}\right]$

$i=\frac{5}{12} \pi J_{0} R^{2}$

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