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(D) The radius at the left end is $a$ and at the right end is $b$. Thus,the increase in radius over the length $\ell$ is $(b - a)$.The rate of increas

Vedclass · Accepted Answer

$\frac{i}{\pi \left[ a + \frac{x(b - a)}{\ell} \right]^2}$

Vedclass · Answer

(D) The radius at the left end is $a$ and at the right end is $b$. Thus,the increase in radius over the length $\ell$ is $(b - a)$.The rate of increase of radius per unit length is $\frac{b - a}{\ell}$.The increase in radius at a distance $x$ from the left end is $\left( \frac{b - a}{\ell} \right) x$.Therefore,the radius $r$ at distance $x$ is $r = a + \left( \frac{b - a}{\ell} \right) x$.The cross-sectional area at this point is $A = \pi r^2 = \pi \left[ a + \left( \frac{b - a}{\ell} \right) x \right]^2$.Since the current $i$ is constant throughout the conductor,the current density $J$ is given by $J = \frac{i}{A} = \frac{i}{\pi \left[ a + \frac{x(b - a)}{\ell} \right]^2}$.

$A$ conductor of length $\ell$ with a circular cross-section is shown in the figure,carrying a current $i$. The radius of the cross-section varies linearly from $a$ to $b$. Assuming $(b - a) << \ell$,calculate the current density at a distance $x$ from the left end.

Similar Questions

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