question_answer

Direction: Question No. 29 are based on the following paragraph. A wire of length L, mass m and carrying a current is suspended from point O as shown. An infinitely long wire carrying the same current I is at a distance L below the lower end of the wire. Given, I = 2A, L= 1m and m = 0.1 kg (ln 2 = 0.693) What is angular acceleration of the wire just after it is released from the position shown?

A)  \[6.2\times {{10}^{-8}}\frac{\text{rad}}{{{\text{s}}^{\text{2}}}}\]            

B)  \[2.1\times {{10}^{-4}}\frac{\text{rad}}{{{\text{s}}^{\text{2}}}}\]

C)  \[4.5\times {{10}^{-5}}\frac{\text{rad}}{{{\text{s}}^{\text{2}}}}\]                            

D)  \[9.3\times {{10}^{-6}}\frac{\text{rad}}{{{\text{s}}^{\text{2}}}}\]

Correct Answer: D

Solution :

\[{{\tau }_{0}}\int\limits_{x=L}^{x=2L}{({{E}_{m}})(2L-x)}\] \[=\int\limits_{L}^{2L}{\left( \frac{{{\mu }_{0}}i}{2\pi x} \right)(i)(2L-x)}dx\] \[=\frac{{{\mu }_{0}}{{i}^{2}}}{2\pi x}[2L\ln x-x]_{L}^{2L}\] \[{{\tau }_{0}}=\frac{{{\mu }_{0}}{{i}^{2}}}{2\pi }[2L\ln (2)-L]\] \[=\frac{(0.386){{\mu }_{0}}{{i}^{2}}L}{2\pi }\]                                    ? (i) \[=(0.386)(2\times {{10}^{-7}})(2)\] \[=3.1\times {{10}^{-7}}N-m\] \[{{I}_{0}}=\frac{m{{L}^{2}}}{3}\] \[=\frac{(0.1){{(1)}^{2}}}{3}=\frac{1}{30}\text{kg-}{{\text{m}}^{\text{2}}}\] \[\therefore \] \[\alpha =\frac{{{\lambda }_{0}}}{{{I}_{0}}}=9.3\times {{10}^{-6}}\text{rad/}{{\text{s}}^{\text{2}}}\]