Question

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

• A. $0.1s$
• B. $0.089s$
• C. $0.076s$
• D. $0.057s$

Answer 1

The correct option is C $0.076s$

Initially $T=2\pi \sqrt{\frac{I}{m{B}_H}},$ finally $T^{\prime }=2\pi \sqrt{\frac{I}{m\left(B\right)}}$

Where $B$ = Magnetic field due to down ward conductor

$B_H=24\times {10}^{–6}T,T_1=0.1s$

$B=B_H-B_{wire}=24\times {10}^{–6}-\frac{{\mu }_{o}i}{2\pi R}=24\times {10}^{–6}-\frac{2\times {10}^{-7}\times 18}{0.2}=14\times {10}^{-6}T$

$\therefore \frac{T^{\prime }}{T}=\sqrt{\frac{B}{B_H}}\Rightarrow \frac{T^{\prime }}{0.1}=\sqrt{\frac{14}{24}}\Rightarrow T^{\prime }=0.076s$