A hollow cylinder has length l,radius r,and t | vedclass

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(B) The hollow cylinder acts as an $L-R$ circuit where the current decays due to its own resistance and inductance.The resistance $R$ of the cylinder

Vedclass · Accepted Answer

$I = I_0 e^{-\frac{2\rho}{\mu_0 r d} t}$

Vedclass · Answer

(B) The hollow cylinder acts as an $L-R$ circuit where the current decays due to its own resistance and inductance.The resistance $R$ of the cylinder for a tangential current path is given by $R = ho \frac{	ext{length of path}}{	ext{cross-sectional area}} = ho \frac{2\pi r}{l d}$.The self-inductance $L$ of a long solenoid (which this cylinder approximates) is $L = \mu_0 n^2 A l$,where $n = \frac{N}{l}$. For a single turn $(N=1)$,$L = \mu_0 \frac{1}{l^2} (\pi r^2) l = \frac{\mu_0 \pi r^2}{l}$.The time constant $	au$ for an $L-R$ circuit is $	au = \frac{L}{R}$.Substituting the values:$	au = \frac{\mu_0 \pi r^2 / l}{ho (2\pi r) / (l d)} = \frac{\mu_0 \pi r^2}{l} \cdot \frac{l d}{2\pi r ho} = \frac{\mu_0 r d}{2ho}$.The current decay is given by $I(t) = I_0 e^{-t/	au}$.Therefore,$I(t) = I_0 e^{-\frac{2ho}{\mu_0 r d} t}$.Comparing this with the given options,the correct expression is $I = I_0 e^{-\frac{2ho}{\mu_0 r d} t}$.

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