Physics - Magnetism Question with Solution | TestHub

The time period of oscillation of a magnetic needle in a uniform magnetic field is given by the formula:

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

Substituting the given values:

$T=2\pi \sqrt{\frac{9.6\times 10^{-5}}{6\times 10^{-2}\times 0.01}}$

$T=2\pi \sqrt{\frac{9.6\times 10^{-5}}{6\times 10^{-4}}}$

$T=2\pi \sqrt{0.16}$

$T=2\pi \times 0.4=0.8\pi \text{ s}$

The time taken to complete 10 oscillations is:

$t=10T=10\times 0.8\pi =8\pi \text{ s}$

Using $\pi =3.14$:

$t=8\times 3.14=25.12\text{ s}$

Answer: 25.12 s