JEE Advanced 2016Paper 1
- Exam
- JEE Advanced
- Year
- 2016
- Paper
- Paper 1
- Subject
- Physics
- Chapter
- Current Electricity
JEE Advanced 2016 · Paper 1 · Physics · Current Electricity
An infinite line charge of uniform electric charge density λ lies...
An infinite line charge of uniform electric charge density lies along the axis of an electrically conducting infinite cylindrical shell of radius R. At time t = 0, the space inside the cylinder is filled with a material of permittivity and electrical conductivity . The electrical conduction in the material follows Ohm's law.
Which one of the following graphs best describes the subsequent variation of the magnitude of current density j(t) at any point in the material?
Which one of the following graphs best describes the subsequent variation of the magnitude of current density j(t) at any point in the material?
Options
- A
- B
- C
- D
Correct answer
A.
Step-by-step solution
This is the problem of RC circuit where the product RC is a constant.
So, due to leakage current, charge and current density will exponentially decay and will infinite time. So correct answer is A
For any small element
Resistance R = \dfrac{dr}{σ\left(2πrl\right)}
Capacitance C = \dfrac{ϵ2πrl}{dr}
Product R × C = \dfrac{ϵ}{σ} = constant
q = {q}_{0}{e}^{- \left(\dfrac{tσ}{ϵ}\right)}
I = \dfrac{dq}{dt} = \dfrac{{q}_{0}σ}{ϵ}{e}^{- \left(\dfrac{tσ}{ϵ}\right)}
Current density = \dfrac{I}{A} = \dfrac{{q}_{0}\dfrac{σ}{ϵ}{e}^{- \dfrac{tσ}{ϵ}}}{2πrl}
j\infty{e}^{- \dfrac{tσ}{ϵ}}
So, due to leakage current, charge and current density will exponentially decay and will infinite time. So correct answer is A
For any small element
Resistance R = \dfrac{dr}{σ\left(2πrl\right)}
Capacitance C = \dfrac{ϵ2πrl}{dr}
Product R × C = \dfrac{ϵ}{σ} = constant
q = {q}_{0}{e}^{- \left(\dfrac{tσ}{ϵ}\right)}
I = \dfrac{dq}{dt} = \dfrac{{q}_{0}σ}{ϵ}{e}^{- \left(\dfrac{tσ}{ϵ}\right)}
Current density = \dfrac{I}{A} = \dfrac{{q}_{0}\dfrac{σ}{ϵ}{e}^{- \dfrac{tσ}{ϵ}}}{2πrl}
j\infty{e}^{- \dfrac{tσ}{ϵ}}