Q. A magnet makes 5 oscillations per min in $B = 0.3 \times 10^{- 4}\ $T. By what amount should the field be increased so that number of oscillations is 10 in the same time?
A) $0.3 \times 10^{- 4}\ \ $T
B) 0.6 $\times 10^{- 4}\ $T
C) 0.9 $\times 10^{- 4\ \ }$T
D) 1.2 $\times 10^{- 4}\ $T
Explanation
(c) $\frac{B_{2}}{B_{1}} = \frac{n_{2}^{2}}{n_{1}^{2}} = \frac{10^{2}}{5^{2}} = 4$
$B_{2} = 4B_{1} = 4 \times 0.3 \times 10^{- 4}T = 1.2 \times 10^{- 4}T$
Increase in field $\ = B_{2} - B_{1} = 0.9 \times 10^{- 4}$ T
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.