JEE MAIN - Physics (2019 - 9th January Morning Slot - No. 4)
Drift speed of electrons, when 1.5 A of current flows in a copper wire of cross section 5 mm2, is $$\upsilon $$. If the electron density in copper is 9 $$ \times $$ 1028/m3 the value of $$\upsilon $$. in mm/s is close to (Take charge of electron to be = 1.6 $$ \times $$ 10$$-$$19C)
0.02
3
2
0.2
Explanation
We know,
I = neAVd
$$ \therefore $$ Vd = $${{\rm I} \over {neA}}$$
= $${{1.5} \over {9 \times {{10}^{28}} \times 1.6{ \times ^{ - 19}} \times 5 \times {{10}^{ - 6}}}}$$
= 0.02 m/s
I = neAVd
$$ \therefore $$ Vd = $${{\rm I} \over {neA}}$$
= $${{1.5} \over {9 \times {{10}^{28}} \times 1.6{ \times ^{ - 19}} \times 5 \times {{10}^{ - 6}}}}$$
= 0.02 m/s
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