Question

A magnet makes $10$ oscillations per minute at a place where the angle of dip is ${45}^0$ and the total intensity is $0.4$ gauss. The number of oscillations made per sec by the same magnet at another place where the angle of dip is ${60}^o$ and the total intensity is $0.5$ gauss is approximately

• A. $6Hz$
• B. $\frac{1}{1.06\times 6}Hz$
• C. $6\times 1.06Hz$
• D. $\frac{1}{6}Hz$

Answer 1

The correct option is B $\frac{1}{1.06\times 6}Hz$

Time period of oscillating magnet is

$T=2\pi \sqrt{\frac{I}{MH}}$

where $H$ is the horizontal component of earth's magnetic field at the place.

At the first place, $H_1=I_1\cos {\varphi }_1=0.4\cos {45}^{\circ }$

At the second place, $H_2=I_2\cos {\varphi }_2=0.5\cos {60}^{\circ }$

where $I_1,I_2$ are the total intensities of earth's magnetic field at two places.

$⟹\frac{T_1}{T_2}=\sqrt{\frac{H_2}{H_1}}$

$⟹T_2=\sqrt{\frac{0.4\cos {45}^{\circ }}{0.5\cos {60}^{\circ }}}\times 6$

Therefore number of oscillations per second=$\frac{1}{T_2}=\frac{1}{1.06\times 6}$