question_answer

A short magnet oscillates with a time period 0.1 s at a place where horizontal magnetic field is \[24\mu T.\] A downward current of 18 A is established in a vertical wire 20 cm east of the magnet. The new time period of oscillator

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 \[{T}'=0.076\,s.\]