Question

When the intensity of magnetic field is increased four times, the time-period of suspended magnetic needle becomes:

• A. double
• B. half
• C. four times
• D. one-fourth less

Answer 1

The correct option is B half

Given: $B^{{}^{\prime }}=4B$

Solution: Formula of time period of suspended magnetic needle is given by:

$T=2\pi \left(\sqrt{\frac{I}{MB}}\right)$

$\frac{T^{{}^{\prime }}}{T}=\sqrt{\frac{B}{B^{{}^{\prime }}}}=\sqrt{\frac{B}{4B}}=\frac{1}{2}$

So, $T^{{}^{\prime }}=\frac{T}{2}$

The time-period of suspended magnetic needle becomes half.

Hence the correct option is B