JEE PHYSICS MCQ: CURRENT ELECTRICITY

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Application +4 | -1

The current density of 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

r=\frac{R}{4}

will be:

View answer and explanation

Correct option: B

\frac{5 J_0 \pi R^2}{96}

Option B (

\frac{5 J_0 \pi R^2}{96}

) is correct because it is the result of integrating the current density function over the specified radial region.

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