A vibration magnetometer consists of two identical bar magnets placed one over the other such that they are…
A vibration magnetometer consists of two identical bar magnets placed one over the other such that they are perpendicular and bisect each other. The time period of oscillation in a horizontal magnetic field $2^{5 / 4} \mathrm{~s}$. One of the magnets is removed and if the other magnet oscillates in the same field, then the time period in seconds is :
$2^{1 / 4}$
$2^{1 / 2}$
2
$2^{5 / 4}$
Solution
Time period of magnet in vibration magnetometer,
$T=2 \pi \sqrt{\frac{I}{M H}}$
First case, $T_1=2 \pi \sqrt{\frac{I_1+I_2}{M^{\prime} H}}$
where, $M=$ resultant magnetic moment of two magnets
Here, two identical magnets are placed perpendicular to each other.
$\therefore \quad I_1=I_2=I$ (Let)
and $\quad M^{\prime}=\sqrt{M^2+M^2}=M \sqrt{2}$
$\therefore \quad T_1=2 \pi \sqrt{\frac{2 I}{\sqrt{2} M H}}$
$2^{5 / 4}=2 \pi \sqrt{\frac{2 I}{\sqrt{2} M H}}$ $\ldots$ (i)
When one magnet is removed, then time period,
$T_2=2 \pi \sqrt{\frac{I}{M H}}$ ...(ii)
On dividing Eq. (i) by Eq. (ii), we get
$\frac{2^{5 / 4}}{T_2}=\sqrt{\frac{2}{\sqrt{2}}}=\sqrt{\sqrt{2}}$
$\frac{2^{5 / 4}}{T_2}=2^{1 / 4}$
$T_2=\frac{2^{5 / 4}}{2^{1 / 4}}=2 \mathrm{~s}$