Magnetic field of the earth is H =0.3 g. A magnet is vibrating 5 oscillations per min then the dippreciation required in the magnetic field of the earth of increase time period upto 10 oscillations per minute is

2.25 g

0.6 g

0.9 g

0.12 g

Solution

2.25 g

The dipole in a uniform magnetic field

T = $2 π \sqrt{\frac{I}{mH}}$

Where I is the moment of inertia

H is the magnetic field

We know that

T $\propto \frac{1}{\sqrt{H}}$

∴ $\frac{T_{1}}{T_{2}} = \sqrt{\frac{H_{2}}{H_{1}}}$

$\frac{5}{10} = \sqrt{\frac{H_{2}}{3}}$

⇒ $\frac{1}{2} = \sqrt{\frac{H_{2}}{3}}$

$\frac{1}{4} = \frac{H_{2}}{3}$

∴ H₂ = 0.75

Dippreciation in field

= 3 - 0.75

= 2.25 g

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