A) \[{{10}^{-2}}\,V/m\]
B) \[{{10}^{-4}}\,V/m\]
C) \[0.1\text{ }V/m\]
D) \[10\text{ }V/m\]
Correct Answer: A
Solution :
| [a] Key Idea: Potential gradient of a wire is equal to potential fall per unit length. |
| Potential gradient = Potential fall per unit length |
| = Current \[\times \] Resistance per unit length |
| \[=i\times \frac{R}{l}\] |
| but \[R=\frac{\rho l}{A}\] |
| \[\Rightarrow \] \[\frac{R}{l}=\frac{\rho }{A}\] |
| \[\therefore \] \[\text{Potential}\,\text{graident}\,=i\times \frac{\rho }{A}\] |
| Here, \[\rho ={{10}^{-7}}\,\Omega -m,\,\,i=0.1\,A,\,A={{10}^{-6}}\,{{m}^{2}}\] |
| Hence, potential gradient \[=0.1\times \frac{{{10}^{-7}}}{{{10}^{-6}}}=\frac{0.1}{10}\] |
| \[=0.01={{10}^{-2}}\,V/m\] |
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