A) 0.2 A
B) 0.6 A
C) 0.8 A
D) 1.2 A
Correct Answer: A
Solution :
The rate of change of flux gives the included emf. Thus, \[{{Q}_{1}}+(-{{W}_{1}})={{Q}_{2}}+(-{{W}_{2}})={{U}_{B}}-{{U}_{A}}\] \[{{F}_{net}}=\frac{{{q}^{2}}\sqrt{2}}{4\pi {{\varepsilon }_{0}}{{a}^{2}}}+\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}{{(\sqrt{2}a)}^{2}}}\] As \[{{F}_{net}}=\left( \frac{1+2\sqrt{2}}{2} \right)\,\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}{{a}^{2}}}\] \[l=\frac{V}{R}=\frac{9}{9}\,=1\,A\] \[R=9\,\Omega \] Induced current \[l=\frac{E}{R}=\frac{2}{10}=0.2\,A\]
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