A) zero
B) 0.66 A towards left
C) 1A towards right I
D) 0.66A towards right
Correct Answer: D
Solution :
Net flow of charge per second is given by \[=(2.9\times {{10}^{18}}+1.2\times {{10}^{18}})\times e\] \[=4.1\times {{10}^{18}}\times 1.6\times {{10}^{-19}}=0.656A\] \[\approx \] \[0.66A\] From this we conclude that the direction of current will be towards right as positive carriers are moving right wards.You need to login to perform this action.
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