A) 0.1 s
B) 0.089 s
C) 0.076 s
D) 0.057 s
Correct Answer: C
Solution :
Initially \[T=2\pi \sqrt{\frac{I}{m{{B}_{H}}}}\] , Finally \[{T}'=2\pi \sqrt{\frac{I}{m(B+{{B}_{H}})}}\] Where B = Magnetic field due to down ward conductor \[=\frac{{{\mu }_{0}}}{4\pi }.\frac{2i}{a}=18\mu T\] \ \[\frac{{{T}'}}{T}=\sqrt{\frac{{{B}_{H}}}{B+{{B}_{H}}}}\] Þ \[\frac{{{T}'}}{0.1}=\frac{24}{18+24}\]Þ \[{T}'=0.076\,s.\]You need to login to perform this action.
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