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

Question

Dr. Abubacker · Accepted Answer

(c) Initially $T = 2\pi\sqrt{\frac{I}{mB_{H}}},\ $Finally $T^{'} = 2\pi\sqrt{\frac{I}{m(B + B_{H})}}$
Where $B =$ Magnetic field due to downward conductor
$= \frac{\mu_{0}}{4\pi}.\frac{2i}{a} = 18\mu T$
$	herefore\frac{T^{'}}{T} = \sqrt{\frac{B_{H}}{B + B_{H}}} \Rightarrow \frac{T^{'}}{0.1} = \frac{24}{18 + 24} \Rightarrow T^{'} = 0.076s$

Q. 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

Explanation

Reviewed by Dr. Abubacker

Want more practice?