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\Delta V measured between B and C

 

Directions : Consider a block of conducting material of resistivity  \rho shown in the figure. Current I enters at A    and leaves from  . We apply superposition principle to find voltage \Delta V developed between  B and C The calculation is done in the following steps:

(i)    Take current  I   entering from  A  and assume it to spread over a hemispherical surface in the block.

(ii)     Calculate field E(r)  at distance r   from  A  by using Ohm’s law E=\rho j  Where J is the current per unit area at r

(iii)     From the r  dependence of E (r) obtain the potential  V (r) at r 

(iv)     Repeat (i),(ii) and (iii) for current  I leaving D and superpose results for A and B

 

 

  • Option 1)

    \frac{\rho I}{2\pi \left ( a-b \right )}

  • Option 2)

    \frac{\rho I}{\pi a}-\frac{\rho I}{\pi \left ( a+b \right )}

  • Option 3)

    \frac{\rho I}{ a}-\frac{\rho I}{ \left ( a+b \right )}

  • Option 4)

    \frac{\rho I}{2\pi a}-\frac{\rho I}{2\pi \left ( a+b \right )}

 

Answers (2)

best_answer

As we learnt in 

If current density is uniform -

\vec{J}=\frac{\vec{I}}{A}

- wherein

\vec{J-} Current density

\vec{A-} Normal cross-section

 

 

 

Current is spread over an area    2\pi r^{2}   The current  I   is a surface current.

Current density, j= \frac{I}{2\pi r^{2}}'

Resistance= \frac{\rho I}{area}= \frac{\rho r}{2\pi r^{2}}

E= \frac{I\rho }{2\pi r^{2}}

V_{B}-V_{C}= \Delta V= \int_{a+b}^{a}-Edr

\Rightarrow \Delta V= \frac{-I\rho }{2\pi }\int_{a+b}^{a}\frac{1}{r^{2}}dr= \frac{-I\rho }{2\pi }\left [ -\frac{1}{r} \right ]_{a+b}^{a}

\Rightarrow \Delta V= \frac{I\rho }{2\pi }\left [ \frac{1}{a}-\frac{1}{a+b} \right ]


Option 1)

\frac{\rho I}{2\pi \left ( a-b \right )}

This option is incorrect.

Option 2)

\frac{\rho I}{\pi a}-\frac{\rho I}{\pi \left ( a+b \right )}

This option is incorrect.

Option 3)

\frac{\rho I}{ a}-\frac{\rho I}{ \left ( a+b \right )}

This option is incorrect.

Option 4)

\frac{\rho I}{2\pi a}-\frac{\rho I}{2\pi \left ( a+b \right )}

This option is correct.

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