Question

A bar magnet, held horizontally, is set into angular oscillations in the earth's magnetic field. Its time periods are $T_1$ and $T_2$ at two places where the angles of dip are ${\theta }_1$ and ${\theta }_2$ respectively. The ratio of the resultant magnetic fields at these two places will be:

• A. $T_{1}sin{\theta }_1:T_{2}sin{\theta }_2$
• B. $T_{1}cos{\theta }_1:T_{2}cos{\theta }_2$
• C. $T_2^{2}sin{\theta }_2:T_1^{2}sin{\theta }_1$
• D. $T_2^{2}cos{\theta }_2:T_1^{2}cos{\theta }_1$

Answer 1

Answer: (A), (D)

The correct options are
A $T_{1}sin{\theta }_1:T_{2}sin{\theta }_2$
D $T_2^{2}cos{\theta }_2:T_1^{2}cos{\theta }_1$

$T_1=2\pi \sqrt{\frac{1}{M{B}_{1}cos{\theta }_1}}$

$T_2=2\pi \sqrt{\frac{1}{M{B}_{2}cos{\theta }_2}}$

$\frac{T_1}{T_2}=\sqrt{\frac{B_{2}cos{\theta }_2}{B_{1}cos{\theta }_1}}$

$\frac{B_1}{B_2}.\frac{cos{\theta }_1}{cos{\theta }_2}=\frac{T_2^2}{T_1^2}$

$\frac{B_1}{B_2}=\frac{T_2^{2}cos{\theta }_2}{T_1^{2}cos{\theta }_1}$