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 :
Solution
<p>$T \propto {1 \over {\sqrt {B\cos \delta } }}$</p>
<p>$$ \Rightarrow {{{T_1}} \over {{T_2}}} = \sqrt {{{{B_2}\cos {\delta _2}} \over {{B_1}\cos {\delta _1}}}} $$</p>
<p>$$ \Rightarrow {{3\,s} \over {6\,s}} = \sqrt {{{{B_1}} \over {{B_2}}} \times {{{1 \over 2}} \over {{{\sqrt 3 } \over 2}}}} $$</p>
<p>$$ \Rightarrow {{{B_2}} \over {{B_1}}} = {\left( {{1 \over 2}} \right)^2} \times \sqrt 3 = {{\sqrt 3 } \over 4}$$</p>
About this question
Subject: Physics · Chapter: Magnetism · Topic: Bar Magnet and Magnetic Field
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