A) \[40\times {{10}^{-8}}N-\sec \]
B) \[30\times {{10}^{-8}}N-\sec \]
C) \[50\times {{10}^{-8}}N-\sec \]
D) \[70\times {{10}^{-8}}N-\sec \]
Correct Answer: A
Solution :
[a] No. of electron in the wire \[=nA\ell \,(n={{e}^{-}}\text{density})\] and momentum \[=n\,eA\,{{v}_{d}}=\frac{\ell {{m}_{e}}}{e}=\frac{I\,\ell {{m}_{e}}}{e}\] \[=\frac{70\times 1000}{1.6\times {{10}^{-19}}}\times 9.1\times {{10}^{-31}}=40\times {{10}^{-8}}N-\sec \]
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