Question

Time period of a bar magnet oscillating in horizontal component of earth's magnetic field in a vibration magnetometer is $3sec$. When another magnet is brought near and parallel to the first, the time period reduces to $1sec$. Then the ratio of magnetic field of the second magnet at the location of first magnet to the horizontal component of earth's field is

• A. $9$
• B. $6$
• C. $10$
• D. $1$

Answer 1

Answer: (A)

$9$

Given: Time period in case 1,$T_1=3Sec$

Time period in case 2,$T_2=1Sec$

To Find: ratio of magnetic field of case 2 to 1 is, $\frac{B_{H2}}{B_{H1}}=?$

Solution: As we all know that in vibration magnetometer

Time period is,

$T=2\pi \sqrt{\frac{I}{M{B}_H}}$ $.....\left(1\right)$

here,$I=$ moment of inertia, $M=$ magnetic moment

on using $e{q}^{n}1$, we get

$T\propto \frac{1}{\sqrt{B_H}}$

$==>B_H\propto \frac{1}{T^2}$

So, $\frac{B_{H2}}{B_{H1}}=(\frac{T_1}{T_2}{)}^2$

$\frac{B_{H2}}{B_{H1}}=(\frac{3}{1}{)}^2$

$==>\frac{B_{H2}}{B_{H1}}=9$

hence,

The correct opt:A