Question

A short magnet oscillates in Vibration Magnetometer with a time period of $0.1sec$ where $B_H$ is $24\mu T$. An upward current of $18A$ is established in the vertical wire placed $20cm$ east of the magnet. The new time period is :

• A. $0.4sec$
• B. $0.2sec$
• C. $0.1sec$
• D. $0.3sec$

Answer 1

Answer: (B)

$0.2sec$

For a Vibration Magnetometer, $T\propto \frac{1}{\sqrt{B_H}}$

The magnetic field due to the current carrying wire will be in the direction shown in the figure. The magnetic field due to the wire will be

$B_w=\frac{{\mu }_{0}i}{2\pi d}=\frac{4\pi \times {10}^{-7}\times 18}{2\pi \times 0.2}=18\mu T$

Given $T_1=0.1s,B_1=B_H=24\mu T$

From the given data $B_2=24-18=6\mu T$

$\therefore \frac{T_2}{T_1}=\sqrt{\frac{B_1}{B_2}}=\sqrt{\frac{24}{6}}=2$

$\therefore T_2=2\times 0.1=0.2s$