Question

A bar magnet is freely oscillating in earth's magnetic field. The magnet is heated so as to reduce its magnetic moment by $19\%$. By doing this, the periodic time of the magnet will,

• A. $\text{Increase by 19\%}$
• B. $\text{Decrease by 19\%}$
• C. $\text{Increase by 11\%}$
• D. $\text{Decrease by 21\%}$

Answer 1

The correct option is C $\text{Increase by 11\%}$

For an oscillating magnet,

$\text{T}=2\pi \sqrt{\frac{I}{M{B}_H}}$

${\text{B}}_H\to \text{Horizontal component of earth's magnetic filed}$

$\Rightarrow \frac{T_1}{T_2}=\sqrt{\frac{M_2}{M_1}}$

Let, $M_1=M\Rightarrow M_2=M-0.19M=0.81M$

$\Rightarrow \frac{T_1}{T_2}=\sqrt{\frac{0.81M}{M}}=0.9$

$\therefore T_2=1.11T_1=T_1+0.11T_1=T_1+11\%\text{of}{T}_1$

So, the time period increases by $11\%$

Hence, $\left(C\right)$ is the correct answer.