2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held On 09-Jan-2019 Morning)

  • question_answer
    Drift speed of electrons, when \[1.5\text{ }A\] of current flows in a copper wire of cross section \[5\text{ }m{{m}^{2}}\], is v. If electron density in copper is \[9\times {{10}^{28}}/{{m}^{3}}\] the value of v in mm/s is close to (Take charge of electron to be \[=\text{ }1.6\times {{10}^{-19}}C)\] [JEE Main Online Paper (Held On 09-Jan-2019 Morning]

    A) 0.02     

    B) 0.2

    C) 3                     

    D)      2

    Correct Answer: A

    Solution :

    Since \[I\text{ }=\text{ }n\text{ }e\text{ }A\text{ }Vd\] \[1.5\text{ }=\text{ }9\text{ }\times \text{ }{{10}^{28}}\times \text{ }1.6\text{ }\times \text{ }{{10}^{-19}}\,\times \text{ }5\text{ }\times \text{ }l{{0}^{-6}}\,V\] \[\Rightarrow \,\,\,V\,\,=\,\,\frac{1.5}{9\,\,\times \,\,1.6\,\,\times \,\,5\,\,\times \,\,{{10}^{3}}}\,\,m/s\] \[V\,\,\,=\,\,\,\frac{0.3}{9\,\,\times \,\,1.6}\,\,mm/s\] \[V\,\,\,=\,\,\frac{1}{48}\,mm/s\] \[=\text{ }0.02\,\,mm/s\] Option is correct.


You need to login to perform this action.
You will be redirected in 3 sec spinner