Question

A short magnet makes $40$ oscillations per minute at a place where horizontal component of the earth's field is $25\mu T$. Another short magnet of dipole moment $1.6A{m}^2$ is placed $20cm$ east of the oscillating magnet. Find the new frequency of oscillation if its $N-$pole is towards north.

• A. $20$
• B. $22.3$
• C. $18$
• D. $54$

Answer 1

The correct option is C $18$

$B=\frac{{\mu }_{o}M}{4\pi d^3}=\frac{4\pi \times {10}^{-7}\times 1.6}{4\pi \times (0.2{)}^3}=20\mu T$ in the opposite direction of $B_H$ as illustrated in Figure.
$\frac{f_1}{f_2}=\frac{T_2}{T_1}=\sqrt{\frac{B_H}{B_H-B}}$.

Thus $f_2=f_1\sqrt{\frac{B_H-B}{B_H}}$
$=40\sqrt{\frac{25-20}{25}}=\frac{40}{2.23}=18$