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AIIMS 201925 May
Exam
AIIMS
Year
2019
Date
25 May
Subject
Physics
Chapter
Current Electricity

AIIMS 2019 · 25 May · Physics · Current Electricity

The current density in a solid cylindrical wire of radius R , as a...

The current density in a solid cylindrical wire of radius RR, as a function of radial distance rr is given by J(r)=J0(1−rR)J(r)=J_0\left(1-\frac{r}{R}\right). The total current in the radial region r=0r=0 to r=R4r=\frac{R}{4} will be

Options

  1. A5J0πR232\frac{5 J_0 \pi R^2}{32}
  2. B5J0πR296\frac{5 J_0 \pi R^2}{96}
  3. C3J0πR294\frac{3 J_0 \pi R^2}{94}
  4. DJ0πR2128\frac{J_0 \pi R^2}{128}

Correct answer

B. 5J0πR296\frac{5 J_0 \pi R^2}{96}

Step-by-step solution

 :di=JdA=J0(1−rR)2πrdri=∫r=0r=R4J0(1−rR)2πrdr=J05πR296\begin{aligned} \text{ } & : d i=J d A=J_0\left(1-\frac{r}{R}\right) 2 \pi r d r \\ 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}\end{aligned}
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