JEE MAIN - Physics (2021 - 27th July Morning Shift - No. 1)
In the given figure, a battery of emf E is connected across a conductor PQ of length 'l' and different area of cross-sections having radii r1 and r2 (r2 < r1).
_27th_July_Morning_Shift_en_1_1.png)
Choose the correct option as one moves from P to Q :
Choose the correct option as one moves from P to Q :
Drift velocity of electron increases.
Electric field decreases.
Electron current decreases.
All of these
Explanation
_27th_July_Morning_Shift_en_1_2.png)
Current is constant in conductor
i = constant
Resistance of element $$dR = {{\rho dx} \over {\pi {r^2}}}$$
$$dV = idR = {{i\rho dx} \over {\pi {r^2}}}$$
$$E = {{dV} \over {dx}} = {{i\rho } \over {\pi {r^2}}}$$
& $${V_d} = {{eE\tau } \over m}$$
$$\therefore$$ $${V_d} \propto E$$
$$ \to E \propto {1 \over {{r^2}}}$$
if r decreases, E will increase : Vd will increase
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