Question

A bar magnet suspended freely in a uniform magnetic field is vibrating with a time period of $3$ seconds. If the field strength is increased to $4$ times of the earlier field strength, the time period will be :

• A. $12s$
• B. $6s$
• C. $1.5s$
• D. $0.75s$

Answer 1

Answer: (C)

$1.5s$

Time period of oscillation $T\propto \sqrt{\frac{1}{B}}$
$\frac{T_1}{T_2}=\sqrt{\frac{B_2}{B_1}}=\sqrt{\frac{4B_1}{B_1}}=\sqrt{4}=2$
$\therefore T_2=\frac{T_1}{2}=\frac{3}{2}=1.5s$