Question

A vibration magnetometer placed in magnetic merlian has a small bar magnet. The magnet executes oscillations with a time period of 2s in earth’s horizontal magnetic field of 24muT. When a horizontal field of 18muT is produced opposite to the earth’s field by placing a current carrying wire, the new time period of the magnet will be then
A. 1s
B. 2s
C. 3s
D. 4s

Answer 1

Hint: The time period of the bar magnet in a magnetic field is given by $T=2\pi \sqrt{\dfrac{I}{MB}}$. Write the equation of time period in both the given cases and then divide them to find the value of the new time period in the second case.
Formula used:
$T=2\pi \sqrt{\dfrac{I}{MB}}$

Complete step-by-step answer:
The time period of the bar magnet in a magnetic field is given by $T=2\pi \sqrt{\dfrac{I}{MB}}$,
where T is the time period of the bar magnet, I is the moment of inertia of the bar magnet about the axis passing through its centre, M is the mass of the bar magnet and B is the net magnetic field.
Let the time period of the small bar magnet when the magnetic field ${{B}_{1}}$ is equal to 24muT be ${{T}_{1}}$. Therefore,
${{T}_{1}}=2\pi \sqrt{\dfrac{I}{M{{B}_{1}}}}$ ……. (i).
It is given that ${{T}_{1}}=2s$.
When a horizontal field of 18muT is produced opposite to the earth’s field by placing a current carrying wire, a new magnetic field will be created because the horizontal field of 18muT opposes the erath’’s field of 24muT.
Let the time period of the bar magnet, when a new magnetic field ${{B}_{2}}$ is created be ${{T}_{2}}$.
Hence, ${{T}_{2}}=2\pi \sqrt{\dfrac{I}{M{{B}_{2}}}}$ ….. (ii).
The new magnetic field created will be equal to (24 – 18) muT.
Hence, ${{B}_{2}}=6muT$.
Divide equation (i) and equation (ii).
Therefore, we get
$\dfrac{{{T}_{1}}}{{{T}_{2}}}=\dfrac{2\pi \sqrt{\dfrac{I}{M{{B}_{1}}}}}{2\pi \sqrt{\dfrac{I}{M{{B}_{2}}}}}$
This implies that,
$\dfrac{{{T}_{1}}}{{{T}_{2}}}=\dfrac{\sqrt{{{B}_{2}}}}{\sqrt{{{B}_{1}}}}\Rightarrow {{T}_{2}}={{T}_{1}}\dfrac{\sqrt{{{B}_{1}}}}{\sqrt{{{B}_{2}}}}$
Substitute the values of ${{T}_{1}}$, ${{B}_{1}}$ and ${{B}_{2}}$.
Therefore, we get
${{T}_{2}}=2\times \dfrac{\sqrt{24}}{\sqrt{6}}=2\times \sqrt{\dfrac{24}{6}}=2\times \sqrt{4}=2\times 2=4s$
Therefore, the new time period of the small bar magnet placed in the vibration magnetometer is 4 seconds.

Note: We can note when the net magnetic field decreased, the time period of oscillations of the bar magnet inside the magnetometer increased. This relation helps to know the value of the horizontal magnetic field inside.