Question 37: Magnetism and Matter

Question

A compass needle of oscillation magnetometer oscillates 20 times per minute at a place $$\mathrm{P}$$ of $$\operatorname{dip} 30^{\circ}$$. The number of oscillations per minute become 10 at another place $$\mathrm{Q}$$ of $$60^{\circ}$$ dip. The ratio of the total magnetic field at the two places $$\left(B_{Q}: B_{P}\right)$$ is :

Accepted Answer

$$T \propto {1 \over {\sqrt {B\cos \delta } }}$$
$$ \Rightarrow {{{T_1}} \over {{T_2}}} = \sqrt {{{{B_2}\cos {\delta _2}} \over {{B_1}\cos {\delta _1}}}} $$
$$ \Rightarrow {{3\,s} \over {6\,s}} = \sqrt {{{{B_1}} \over {{B_2}}} 	imes {{{1 \over 2}} \over {{{\sqrt 3 } \over 2}}}} $$
$$ \Rightarrow {{{B_2}} \over {{B_1}}} = {\left( {{1 \over 2}} ight)^2} 	imes \sqrt 3  = {{\sqrt 3 } \over 4}$$

Answer

$$\sqrt{3}: 4$$

Answer

$$4: \sqrt{3}$$

Answer

$$\sqrt{3}: 2$$

Answer

$$2: \sqrt{3}$$

Magnetism and Matter

Choose the correct answer: