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Question

(B) The frequency of oscillation of a magnet in a horizontal plane is given by $
u = \frac{1}{2\pi} \sqrt{\frac{MB_H}{I}}$,where $B_H = B \cos \phi$

Vedclass · Accepted Answer

$16 : 9\sqrt{3}$

Vedclass · Answer

(B) The frequency of oscillation of a magnet in a horizontal plane is given by $
u = \frac{1}{2\pi} \sqrt{\frac{MB_H}{I}}$,where $B_H = B \cos \phi$ is the horizontal component of the Earth's magnetic field,$B$ is the total magnetic field,and $\phi$ is the angle of dip.Given $
u_1 = \frac{20}{60} = \frac{1}{3} 	ext{ Hz}$ at $\phi_1 = 30^{\circ}$ and $
u_2 = \frac{15}{60} = \frac{1}{4} 	ext{ Hz}$ at $\phi_2 = 60^{\circ}$.Since $
u \propto \sqrt{B \cos \phi}$,we have $\frac{
u_1}{
u_2} = \sqrt{\frac{B_1 \cos \phi_1}{B_2 \cos \phi_2}}$.Squaring both sides: $\frac{B_1}{B_2} = \left( \frac{
u_1}{
u_2} ight)^2 \frac{\cos \phi_2}{\cos \phi_1}$.Substituting the values: $\frac{B_1}{B_2} = \left( \frac{1/3}{1/4} ight)^2 	imes \frac{\cos 60^{\circ}}{\cos 30^{\circ}} = \left( \frac{4}{3} ight)^2 	imes \frac{1/2}{\sqrt{3}/2} = \frac{16}{9} 	imes \frac{1}{\sqrt{3}} = \frac{16}{9\sqrt{3}}$.

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