Q. A magnet performs 10 oscillations per minute in a horizontal plane at a place where the angle of dip is 45${^\circ}$ and the total intensity is 0.707 CGS units. The number of oscillations per minute at a place where dip angle is 60${^\circ}$ and total intensity is 0.5 CGS units will be
A) 5
B) 7
C) 9
D) 11
Explanation
(b) Here, $n_{1} = 10$ oscillations per min
$\delta_{1} = 45{^\circ},\ T_{1} = 0.707$ CGS units
$n_{2} = ?,\delta_{2} = 60{^\circ},R_{2} = 0.5$ CGS units
$\ \frac{n_{2}}{n_{1}} = \sqrt{\frac{H_{2}}{H_{1}}} = \sqrt{\frac{R_{2}\cos\delta_{2}}{R_{1}\cos\delta_{1}}}$
$\frac{n_{2}}{10} = \sqrt{{^\circ}\frac{0.5cos60{^\circ}}{0.707cos45{^\circ}} = \ }\sqrt{\frac{0.5 \times 1/2}{0.5 \times \sqrt{2} \times 1/\sqrt{2}} = \frac{1}{\sqrt{2}}}$
$n_{2} = \frac{10}{\sqrt{2}} = 7.07$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.