 The bar magnet
Question

# A magnet oscillating in a  horizontal plane has a time period of 2 sec at a place where the angle of dip is ${30}^{\mathrm{o}}$ and 3 second at another place where the angle of dip is ${60}^{\mathrm{o}}$. The ratio of resultant magnetic fields at these two places is

Moderate
Solution

## $\begin{array}{l}{\left({\mathrm{B}}_{\mathrm{H}}\right)}_{1}={\mathrm{B}}_{1}\mathrm{cos}{30}^{\mathrm{o}}\\ {\left({\mathrm{B}}_{\mathrm{H}}\right)}_{2}={\mathrm{B}}_{2}\mathrm{cos}{60}^{\mathrm{o}}\\ \mathrm{T}\propto \frac{1}{\sqrt{{\mathrm{B}}_{\mathrm{H}}}}\\ \frac{{\mathrm{T}}_{1}}{{\mathrm{T}}_{2}}=\sqrt{\frac{{\mathrm{B}}_{{\mathrm{H}}_{2}}}{{\mathrm{B}}_{{\mathrm{H}}_{1}}}}=\sqrt{\frac{{\mathrm{B}}_{2}\mathrm{Cos}{60}^{\mathrm{o}}}{{\mathrm{B}}_{1}\mathrm{cos}{30}^{\mathrm{o}}}}\\ \frac{2}{3}=\sqrt{\frac{{\mathrm{B}}_{2}}{{\mathrm{B}}_{1}}\frac{1}{\sqrt{3}}}\\ \frac{4}{9}=\frac{{\mathrm{B}}_{2}}{{\mathrm{B}}_{1}}\frac{1}{\sqrt{3}}\\ \frac{{\mathrm{B}}_{1}}{{\mathrm{B}}_{2}}=\frac{9}{4\sqrt{3}}\end{array}$

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