Rankers Physics

Magnetic Properties of Matter: Practice Problem & Solution

14. A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of $2text{ sec}$ in earth's horizontal magnetic field of $24text{ microtesla}$. When a horizontal field of $18text{ microtesla}$ is produced opposite to the earth's field by placing a current carrying wire, the new time period of magnet will be: (2010 Pre)
$4text{ s}$
$1text{ s}$
$2text{ s}$
$3text{ s}$

Solution Explained:

To solve this problem, we apply the core principles of Magnetic Properties of Matter. Understanding the underlying formula is key to arriving at the correct answer below:

Time period is $T propto frac{1}{sqrt{B}}$. Initially $B_1 = 24 mutext{T}$. In the second case, the net field is $B_2 = 24 - 18 = 6 mutext{T}$. Therefore, $T_2 = T_1 sqrt{frac{B_1}{B_2}} = 2 times sqrt{frac{24}{6}} = 2 times 2 = 4text{ s}$.

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