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Standard XII

Physics

Conductivity and Resistivity

Two long coax...

Question

Two long coaxial and conducting cylinders of radius $a$ and $b$ are separated by a material of conductivity $σ$ and a constant potential difference $V$ is maintained between them, by a battery. Then the current, per unit length of the cylinder following from one cylinder the other is:

\frac{4 π σ}{ℓ n (b / a) V}

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\frac{4 π σ}{(b + a) V}

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\frac{2 π σ}{(b + a) V}

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\frac{2 π σ}{ℓ n (b / a) V}

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Solution

The correct option is D $\frac{2 π σ}{ℓ n (b / a) V}$
$I = I / A = σ E .$

$E = \frac{I}{2 π σ L} \frac{1}{s}^r .$

$V = \int_{a}^{b} E \cdot d s = \int_{a}^{b} (\frac{I}{2 π σ L}) \frac{1}{s} d s .$

$= \frac{I}{2 π σ L} ln (b / a)$

$I = \frac{(2 π σ L) V}{ln (b / a)}$

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Two long coaxial and conducting cylinders of radius a and b are separated by a material of conductivity σ and a constant potential difference V is maintained between them, by a battery. Then the current, per unit length of the cylinder following from one cylinder the other is:

The correct option is D 2πσℓn(b/a)VI=I/A=σE. E=I2πσL1s^r. V=∫baE⋅ds=∫ba(I2πσL)1sds. =I2πσLln(b/a) I=(2πσL)Vln(b/a)

Two long coaxial and conducting cylinders of radius $a$ and $b$ are separated by a material of conductivity $σ$ and a constant potential difference $V$ is maintained between them, by a battery. Then the current, per unit length of the cylinder following from one cylinder the other is:

The correct option is D $\frac{2 π σ}{ℓ n (b / a) V}$
$I = I / A = σ E .$

$E = \frac{I}{2 π σ L} \frac{1}{s}^r .$

$V = \int_{a}^{b} E \cdot d s = \int_{a}^{b} (\frac{I}{2 π σ L}) \frac{1}{s} d s .$

$= \frac{I}{2 π σ L} ln (b / a)$

$I = \frac{(2 π σ L) V}{ln (b / a)}$