Current density in a cylindrical wire of radi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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:

Similar Questions

$A$ current flows in a wire of circular cross-section with the free electrons travelling with a mean drift velocity $\vec v$. If an equal current flows in a wire of twice the radius,the new mean drift velocity is:

The drift velocity of free electrons in a conductor is $v$ when a current $i$ is flowing in it. If both the radius and current are doubled,then the drift velocity will be

The drift speed of electrons in a material is found to be $0.3 \ m s^{-1}$ when an electric field of $2 \ V m^{-1}$ is applied across it. The electron mobility (in $m^2 \ V^{-1} \ s^{-1}$) in the material is

$A$ current $I$ flows through a uniform wire of diameter $d$ when the mean electron drift velocity is $V$. The same current will flow through a wire of diameter $d/2$ made of the same material if the mean drift velocity of the electron is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module