Question

The magnetic needle of a vibration magnetometer completes 10 oscillations in 92 seconds. When a small magnet is placed in the magnetic meridian 10cm due north of the needle with north pole towards south completes 15 oscillations in 69 seconds. The magnetic moment of the magnet $(B_H=0.3G)$ is

• A. 4.5 Am${}^2$
• B. 0.45Am${}^2$
• C. 0.75 Am${}^2$
• D. 0.225 Am${}^2$

Answer 1

The correct option is B 0.45Am${}^2$

Let the magnetic field due to the small magnet be $B$ at the site of oscillating needle.As the magnetic field is towards North thus the resultant magnetic field is $B+B_H$

Using, $T_1=2\pi \sqrt{\frac{I}{M{B}_H}}$ and $T_2=2\pi \sqrt{\frac{I}{M(B+B_H)}}$ where $M$ is the magnetic moment of the needle.

Time period, $T_1=\frac{92}{10}=9.2s$ $T_2=\frac{69}{15}=4.6s$

Thus $(\frac{T_1}{T_2}{)}^2=\frac{B+B_H}{B_H}$

Given: $B_H=.3G=.3\times {10}^{-4}T$

$(\frac{9.2}{4.6}{)}^2=\frac{B+.3\times {10}^{-4}}{.3\times {10}^{-4}}$

$⟹B=.9\times {10}^{-4}T$

Now $B=\frac{{\mu }_o}{4\pi }\frac{2M_m}{d^3}$ where$M_m$ is the magnetic moment of the small magnet.

Thus $.9\times {10}^{-4}={10}^{-7}\times \frac{2M_m}{{.1}^3}$

$⟹M_m=.45A{m}^2$