Question

A short magnet oscillates with a time period 0.1 s at a place where horizontal magnetic field is 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

Answer 1

The correct option is C 0.076 s


Initially $T=2\pi \sqrt{\frac{I}{m{B}_H}},Finally{T}^{\prime }=2\pi \sqrt{\frac{I}{m(B+B_H)}}$
Where B = Magnetic field due to down ward conductor
$=\frac{{\mu }_0}{4\pi }.\frac{2i}{a}=18\mu T\therefore \frac{T^{\prime }}{T}=\sqrt{\frac{B_H}{B+B_H}}\Rightarrow \frac{T^{\prime }}{0.1}=\frac{24}{18+24}\Rightarrow T^{\prime }=0.076s.$