Question

Magnetic field of earth is $0.3\text{G}$. A magnet is oscillating with rate of $5\text{oscillations/min}$. By what percentage, should the magnetic field of earth be changed, so that the number of oscillations become $4\text{oscillations/min}$?

• A. Increased by $72\%$
• B. Decreased by $72\%$
• C. Increased by $36\%$
• D. Decreased by $36\%$

Answer 1

The correct option is D Decreased by $36\%$

The time period of vibration of a magnet in a magnetic field is given by,

$T=2\pi \sqrt{\frac{I}{MB}}$

$\Rightarrow f=\frac{1}{T}=\frac{1}{2\pi }\sqrt{\frac{MB}{I}}$

From the above equation, we can say that,

$f\propto \sqrt{B}$

$\Rightarrow \frac{f_1^2}{f_2^2}=\frac{B_1}{B_2}$

$\Rightarrow \frac{5^2}{4^2}=\frac{0.3}{B_2}$

$\Rightarrow B_2=0.192\text{G}$

$\therefore \Delta B=B_1-B_2=0.3-0.192=0.108\text{G}$

Now, percentage change in $B$,

$\frac{\Delta B}{B_1}\times 100=\frac{0.108}{0.3}\times 100=36\%$

As, $B_1>B_2$, the magnetic field should be decreased by $36\%$.

Hence, option $\left(D\right)$ is correct.