A conductor of resistivity ρ and resistance R, as shown in the figure, is connected across a battery of mf V. Its radius varies from 'a' at left end to 'b' at right end. The electric field at a point P at distance x from left end of it is:

Engineering

Physics

Ohm Law and Current Basics

Question

none of these

$\frac{2 V l^{2} ρ}{πR {(l a + (b - a) x)}^{2}}$

$\frac{V l^{2} ρ}{2 πR {(l a + (b - a) x)}^{2}}$

$\frac{V l^{2} ρ}{πR {(l a + (b - a) x)}^{2}}$

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E = jr [j = current density]
j = $\frac{I}{{πr}^{2}}$ [r = radius of c.s. at distance 'x' from left end]
$r = [a + \frac{(b - a)}{l} x]$
Hence, $E = \frac{{Vl}^{2} ρ}{πR (al + (b - a) x)^{2}}$

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