A magnet performs 10 oscillations per minute in a horizontal plane at a place where the angle o dip is $45{^\circ}$ and the total intensity is 0.707 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

Question

Dr. Abubacker · Accepted Answer

(b) Here, $n_{1} = 10\ $oscillation per min
$\delta_{1} = 45{^\circ},\ B_{1} = 0.707\ $CGS units
$n_{2} = ?,\ \delta_{2}60{^\circ},\ B_{2} = 0.5$ CGS units
$\frac{n_{2}}{n_{1}} = \sqrt{\frac{H_{2}}{H_{1}}} = \sqrt{\frac{B_{2}\cos\delta_{2}}{B_{1}\cos\delta_{1}}}$
$\frac{n_{2}}{n_{1}} = \sqrt{\frac{0.5\cos{60{^\circ}}}{0.707\ \cos{45{^\circ}}}}\  = \sqrt{\frac{0.5 	imes 1/2}{0.5 	imes \sqrt{2} 	imes \frac{1}{\sqrt{2}}}}$
$\frac{n_{2}}{10} = \frac{1}{\sqrt{2}} \Longrightarrow n_{2} = \frac{10}{\sqrt{2}} = 7.07 \approx 7$

Q. A magnet performs 10 oscillations per minute in a horizontal plane at a place where the angle o dip is $45{^\circ}$ and the total intensity is 0.707 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

Explanation

Reviewed by Dr. Abubacker

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