Question

The period of a magnet in earth's field $B_0$ is $T$. When another magnet is placed at certain distance on the magnetic meridian in the same line as the centre of the vibrating magnet, the period is $T_1$. When the magnet in the same position is reversed the period is $T_2$. [The field due to the magnet is smaller than the earth's horizontal component]. Then the field due to the magnet at the centre of the vibrating magent?

• A. $\frac{T_2^2-T_1^2}{T_2^2+T_1^2}{B}_0$
• B. $\frac{T_1^2}{T_2^2}{B}_0$
• C. $\frac{T_2^2+T_1^2}{T_2^2+T_1^2}{B}_0$
• D. $B_0\sqrt{\frac{T_2}{T_1}}$

Answer 1

The correct option is A $\frac{T_2^2-T_1^2}{T_2^2+T_1^2}{B}_0$

Using the formula,

Time period of oscillation of magnet is given by

$T=2\pi \sqrt{\frac{I}{MB}}$

Given,

Earth's magnetic field $B_0$

Let the magnetic field due to the magnet at the centre of the vibrating magnet be $B$.

We have,

CASE $1$:

When the magnetic field due to the earth and the another magnet adds up

$B_1=B_0+B$

$\Rightarrow T_1=2\pi \sqrt{\frac{I}{M{B}_1}}=2\pi \sqrt{\frac{I}{M(B_0+B)}}$

CASE $2$:

When the magnet in the same position is reversed

$B_2=B_0-B$

Given, $B_0>B$

$\Rightarrow T_2=2\pi \sqrt{\frac{I}{M{B}_2}}=2\pi \sqrt{\frac{I}{M(B_0-B)}}$

$\frac{T_1}{T_2}=\frac{\left(\sqrt{\frac{I}{M(B_0+B)}}\right)}{\left(\sqrt{\frac{I}{M(B_0-B)}}\right)}$

$\Rightarrow \frac{{T_1}^2}{{T_2}^2}=\frac{B_0-B}{B_0+B}$

Applying Componendo and Dividendo, we have

$\Rightarrow \frac{{T_1}^2+{T_2}^2}{{T_1}^2-{T_2}^2}=\frac{B_0-B+B_0+B}{B_0-B-B_0-B}$

$\Rightarrow \frac{{T_1}^2+{T_2}^2}{{T_1}^2-{T_2}^2}=\frac{2B_0}{-2B}$
$\Rightarrow \frac{{T_2}^2+{T_1}^2}{{T_2}^2-{T_1}^2}=\frac{B_0}{B}$

$\Rightarrow B=B_0\frac{{T_2}^2-{T_1}^2}{{T_2}^2+{T_1}^2}$

Hence, the correct answer is OPTION A.