Question

The period of oscillations of a magnetic needle in a magnetic field is given as $1.0\,\sec $. What will be the time period, if the length of the needle is halved?
A. $1.0\,\sec $
B. $0.5\,\sec $
C. $0.25\,\sec $
D. $2.0\,\sec $

Answer 1

Hint: Start with finding the relationship between the time period of oscillations of a magnetic needle in a magnetic field and the length of the needle. Then put all the values from the information provided in the question and finally you will get the required answer for the question given.

Formula used:
The formula of time period is,
$T = 2\pi \sqrt {\dfrac{I}{{MB}}}$
Here, $I$ is the moment of inertia, $M$ is the magnetic moment and $B$ is the magnetic field.

Complete step by step solution:
We know the formula of time period in terms of length of the needle is given by:
$T = 2\pi \sqrt {\dfrac{I}{{MB}}} \\
\Rightarrow T= 2\pi \sqrt {\dfrac{{w{l^2}/12}}{{\text{pole strength} \times 2l \times B}}} $
Therefore, we get;
$T \propto \sqrt {wl} $
Using this equation we get;
$\dfrac{{{T_1}}}{{{T_2}}} = \sqrt{\dfrac{w_2\,l_2}{w_1\,l_1}} \\ $
$\Rightarrow \dfrac{{{T_1}}}{{{T_2}}} = \sqrt{\dfrac{\dfrac{w_1}{2}\,\dfrac{l_1}{2}}{w_1\,l_1}} \\ $
$\therefore {T_2} = \dfrac{{{T_1}}}{2} = 0.5\,\sec $

Hence the correct answer is option B.

Note: Here the length of the needle is get halved if it get three fourth then the answer will get changed according to the value of the length of the needle given in the question so try to put the values from the question carefully in the required equation in the answer otherwise you will not get the right answer.