A) \[\frac{2{{\mu }_{0}}{{U}_{0}}}{\pi \lambda Rd}\]
B) \[\frac{{{\mu }_{0}}{{U}_{0}}}{2\pi \lambda Rd}\]
C) \[\frac{{{\mu }_{0}}{{U}_{0}}}{\pi \lambda Rd}\]
D) \[\frac{{{\mu }_{0}}{{U}_{0}}}{4\pi \lambda Rd}\]
Correct Answer: B
Solution :
[b] We will assume that the capacitor discharges quickly and there is no appreciable displacement of the wires in that interval. Let initial charge on the capacitor be \[{{Q}_{0}}.\] Current at time t is \[I=\frac{{{Q}_{0}}}{RC}{{e}^{-t/RC}}\] Force between two parallel wires per unit length is \[F=\frac{{{\mu }_{0}}{{I}^{2}}}{2\pi d}=\frac{{{\mu }_{0}}Q_{0}^{2}}{2\pi d{{R}^{2}}{{C}^{2}}}{{e}^{-2t/RC}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\lambda dv=\frac{{{\mu }_{0}}Q_{0}^{2}}{2\pi d{{R}^{2}}{{C}^{2}}}{{e}^{-2t/RC}}dt\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\int\limits_{0}^{v}{dv=\frac{{{\mu }_{0}}Q_{0}^{2}}{2\pi \lambda d{{R}^{2}}{{C}^{2}}}}\int\limits_{0}^{\infty }{{{e}^{-2t/RC}}dt}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,v=\frac{{{\mu }_{0}}Q_{0}^{2}}{4\pi \lambda RCd}=\frac{{{\mu }_{0}}{{U}_{0}}}{2\pi \lambda Rd}\,\,\,\,\,\,\left[ \because \,\,\,{{U}_{0}}=\frac{Q_{0}^{2}}{2C} \right]\]You need to login to perform this action.
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