Question

Two identical bar magnets are placed one above the other such that they are mutually perpendicular and bisect each other. The time period of this combination in a horizontal magnetic field is $T$. The time period of each magnet in the same field is

• A. $\sqrt{2} T$
• B. $2^{\frac{1}{4}} T$
• C. $2^{-\frac{1}{4}} T$
• D. $2^{-\frac{1}{2}} T$

Answer 1

Answer: (C)

Time period of combination,

$T=2 \pi \sqrt{\frac{2 I}{\sqrt{2} M .H}}$ ...(i)
and time period of each magnet,
$T'=2 \pi \sqrt{\frac{T}{M H}}$ ...(ii)
From (i) and (ii), we get
$T'=\frac{T}{2^{1 / 4}}=2^{-1 / 4} T$