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(D) The total current $I$ is given by the integral $I = \int J \cdot dA$. For a cylindrical wire,the area element is $dA = 2\pi x dx$.$I = \int_{0}^{R

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$\frac{5}{12} \pi J_0 R^2$

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(D) The total current $I$ is given by the integral $I = \int J \cdot dA$. For a cylindrical wire,the area element is $dA = 2\pi x dx$.$I = \int_{0}^{R/2} J_0 \left( \frac{x}{R} - 1 \right) (2\pi x dx) + \int_{R/2}^{R} J_0 \frac{x}{R} (2\pi x dx)$$I = 2\pi J_0 \left[ \int_{0}^{R/2} \left( \frac{x^2}{R} - x \right) dx + \int_{R/2}^{R} \frac{x^2}{R} dx \right]$$I = 2\pi J_0 \left[ \left( \frac{x^3}{3R} - \frac{x^2}{2} \right)_{0}^{R/2} + \left( \frac{x^3}{3R} \right)_{R/2}^{R} \right]$$I = 2\pi J_0 \left[ \left( \frac{R^3/8}{3R} - \frac{R^2/4}{2} \right) + \left( \frac{R^3}{3R} - \frac{R^3/8}{3R} \right) \right]$$I = 2\pi J_0 \left[ \left( \frac{R^2}{24} - \frac{R^2}{8} \right) + \left( \frac{R^2}{3} - \frac{R^2}{24} \right) \right]$$I = 2\pi J_0 \left[ \frac{R^2 - 3R^2}{24} + \frac{8R^2 - R^2}{24} \right] = 2\pi J_0 \left[ -\frac{2R^2}{24} + \frac{7R^2}{24} \right]$$I = 2\pi J_0 \left( \frac{5R^2}{24} \right) = \frac{5}{12} \pi J_0 R^2$.

Current density in a cylindrical wire of radius $R$ is given as $J = \begin{cases} J_0 \left( \frac{x}{R} - 1 \right) & 0 \leqslant x < \frac{R}{2} \\ J_0 \frac{x}{R} & \frac{R}{2} \leqslant x \leqslant R \end{cases}$. The current flowing in the wire is:

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