Question

At a certain place, a magnet makes 30 oscillations per min. At another place, where the magnetic field is double, its time period will be:
A. \[4s\]
B. \[2s\]
C. $\dfrac{1}{2}s$
D. $\sqrt{2}s$

Answer 1

Hint: The magnet's time period is equal to the square root of the ratio of its moment of inertia to the sum of its magnetic moment and ambient magnetic field. The magnetic fields for the configurations utilising magnets with various moments will not change. Find the ratio of time periods using this.

Formula used:
\[T=2\pi \sqrt{\dfrac{1}{mB}}\]
where T is the time period of the magnet. m is the magnetic pole strength and B is the magnetic field.
\[T=\dfrac{1}{f}\]
where f is the frequency.

Complete step by step solution:
According to the question, in the first case. A magnet makes 30 oscillations per min so we can say that we are given the frequency of the magnet.
$f=30{{\min }^{-1}}=\dfrac{30}{60}=0.5{{\sec }^{-1}}$
Using the formula \[T=\dfrac{1}{f}\] then ${{T}_{1}}=\dfrac{1}{0.5}=2s$
So now putting the value of ${{T}_{1}}$ in the formula of \[T=2\pi \sqrt{\dfrac{1}{mB}}\] then,
$2=2\pi \sqrt{\dfrac{1}{m{{B}_{1}}}}$………. Equation (1)

Now we have been asked to find the new time period if the magnetic field is doubled then,
${{T}_{2}}=2\pi \sqrt{\dfrac{1}{m{{B}_{2}}}}$………. Equation (2)
Dividing equation 1 by equation 2 we’ll get, \[\dfrac{2}{{{T}_{2}}}=\sqrt{\dfrac{{{B}_{2}}}{{{B}_{1}}}}\]
We need to find the value of ${{B}_{2}}$.

We are given that the magnetic field is getting doubled. Therefore, ${{B}_{2}}=2{{B}_{1}}$ and we need to calculate ${{T}_{2}}$. So, we can write
\[\dfrac{2}{{{T}_{2}}}=\sqrt{\dfrac{{{B}_{2}}}{{{B}_{1}}}}\]
\[\Rightarrow \dfrac{{{(2)}^{2}}}{{{T}_{2}}^{2}}=\dfrac{2{{B}_{1}}}{{{B}_{1}}}\]
$\therefore 2T_{2}^{2}=4$
Or we can say that $T_{2}^{2}=2$ and hence ${{T}_{2}}=\sqrt{2}s$

Hence, the correct option is D.

Notes: By balancing the moment (spinning acceleration) produced by the displacement and the moment due to the magnetic field, you may determine the oscillation's time period. This will create a harmonic oscillation and, as a result, a relationship for time.