The current density is a solid cylindrical wire of radius R , as a function of radial distance r is given by J ( r

Q: The current density is a solid cylindrical wire of radius $R ,$ as a function of radial distance $r$ is given by $J ( r )= J _{0}\left(1-\frac{ r }{ R }\right) .$ The total current in the radial region $r =0$ to $r =\frac{ R }{4}$ will be

b Consider the expression, $d i=J d A$ $=J_{0}\left(1-\frac{r}{R}\right) 2 \pi r d r$ Integrate on both the sides $i=\int_{r=0}^{r=\frac{R}{4}} J_{0}\left(1-\frac{r}{R}\right) 2 \pi r d r$ $=\frac{J_{0} 5 \pi R^{2}}{96}$

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

The current density is a solid cylindrical wire of radius $R ,$ as a function of radial distance $r$ is given by $J ( r )= J _{0}\left(1-\frac{ r }{ R }\right) .$ The total current in the radial region $r =0$ to $r =\frac{ R }{4}$ will be

A$\frac{5 J _{0} \pi R ^{2}}{32}$
B$\frac{5 J _{0} \pi R ^{2}}{96}$
C$\frac{3 J _{0} \pi R ^{2}}{64}$
D$\frac{ J _{0} \pi R ^{2}}{128}$

AIIMS 2019, Diffcult

STD 11 Science
JEE
STD 12 - 3. current electricity
Physics

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