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In a gas discharge tube if 3*10^{18} electrons are flowing per sec from left to right and 2*10^{18} protons are flowing per second from right to left through a given cross section.  The magnitude and direction of current through the cross section

  • Option 1)

    0.48, left to right            

  • Option 2)

    0.48 A, right to left

     

  • Option 3)

    0.80A, left to right

  • Option 4)

    0.80 A, right to left

 

Answers (1)

best_answer

As we learnt

 

Translatory motion of charge -

 

i=\frac{nq}{t}

- wherein

If n particles each having charge q pass a given area in time t

 

 

 

As current is rate of flow of charge in the direction in which positive charge will move, the current due to electron will be
ie = \frac{{{n_e}{q_e}}}{t}$ = 3 ´ 1018 ´ 1.6 ´ 10-19
 = 0.48 A (Opposite to the motion of electrons, i.e. right to left)

                        Current due to protons
            ip =\frac{{{n_p}{q_p}}}{t}$ = 2 ´ 1018 ´ 1.6 ´ 10-19
             = 0.32 A  (Right to left)
so total I = ie + ip = 0.48 + 0.32
            = 0.80 A (Right to left)
 


Option 1)

0.48, left to right            

Option 2)

0.48 A, right to left

 

Option 3)

0.80A, left to right

Option 4)

0.80 A, right to left

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