First slide

Magnetic dipole in a uniform magnetic field

Question

In a vibration magnetometer, the time period of a bar magnet oscillating in horizontal component of earth's magnetic field is 2 sec. When the magnet is brought near a field magnet F and parallel to it, the time period reduces to 1 sec. The ratio of F/H of the horizontal component H and the field F due to magnet will be

Difficult

A

3
B

1/3
C

$\sqrt{3}$
D

$\frac{1}{\sqrt{3}}$

Solution

We know that $T = 2 π \sqrt{(\frac{I}{{MB}_{H}})}$
$Here, T = 2 = 2 π {[\frac{I}{MH}]}^{1 / 2} where B_{H} = H$
When &e magnet is brought near and parallel to it, the time period reduces to 1 sec. i.e., time period decreases. Hence the field is now (H + F). Hence
$\begin{array}{l} T^{'} = 1 = 2 π {[\frac{I}{M (H + F)}]}^{1 / 2} \\ ∴ \frac{2}{1} = {[\frac{H + F}{H}]}^{1 / 2} or \frac{4}{1} = (\frac{H + F}{H}) \\ or 4 = 1 + \frac{F}{H} or \frac{F}{H} = 3 \end{array}$

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