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NEET PHYSICSMedium

A vibration magnetometer placed in a magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of 2 s in earth's horizontal magnetic field of $24\ \mu\text{T}$ . When a horizontal field of $18\ \mu\text{T}$ is produced opposite to the earth's field by placing a current-carrying wire, the new time period of the magnet will be:

A

1 s

B

2 s

C

3 s

D

4 s

Step-by-Step Solution

Time Period Formula: The time period ( $T$ ) of a magnet oscillating in a uniform magnetic field ( $B$ ) is given by $T = 2\pi \sqrt{\frac{I}{mB}}$ , where $I$ is the moment of inertia and $m$ is the magnetic moment. This implies $T \propto \frac{1}{\sqrt{B}}$ .
Initial Condition:

Initial magnetic field ( $B_1$ ) = Earth's horizontal field ( $H$ ) = $24\ \mu\text{T}$ .
Initial time period ( $T_1$ ) = 2 s.

Final Condition:

An external field ( $B_{ext}$ ) of $18\ \mu\text{T}$ is produced opposite to the Earth's field.
Net magnetic field ( $B_2$ ) = $H - B_{ext} = 24\ \mu\text{T} - 18\ \mu\text{T} = 6\ \mu\text{T}$ .

Calculation:

Using the relation $\frac{T_2}{T_1} = \sqrt{\frac{B_1}{B_2}}$ : $\frac{T_2}{2} = \sqrt{\frac{24}{6}} = \sqrt{4} = 2$ $T_2 = 2 \times 2 = 4 \text{ s}$

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