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Question: The area of cross section of a current carrying conductor is A<sub>0</sub> and A<sub>0</sub>/4 at se...

The area of cross section of a current carrying conductor is A0 and A0/4 at section (1) and (2) respectively. If and be the drift velocity at sections (1) and (2) respectively, then –

A

= 1 : 4

B

= 4 : 1

C

= 1 : 1

D

None of these

Answer

= 1 : 4

Explanation

Solution

i = neAvd \ Avd = const. \ vd µ 1 A\frac { 1 } { \mathrm {~A} }

\ vd1vd2=A0/4 A0=14\frac { \mathrm { v } _ { \mathrm { d } _ { 1 } } } { \mathrm { v } _ { \mathrm { d } _ { 2 } } } = \frac { \mathrm { A } _ { 0 } / 4 } { \mathrm {~A} _ { 0 } } = \frac { 1 } { 4 }