A) \[\frac{ml}{N{{e}^{2}}{{A}^{2}}\tau }\]
B) \[\frac{2m\tau A}{N{{e}^{2}}l}\]
C) \[\frac{N{{e}^{2}}\tau A}{2ml}\]
D) \[\frac{N{{e}^{2}}}{2m\tau l}\]
Correct Answer: A
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