Question

A bar magnet of moment of inertia $49\times {10}^{-2}{\text{kg m}}^2$ oscillates in a uniform magnetic field of induction $0.5\times {10}^{-4}\text{T}.$ The time period of oscillation is $8.8\text{s}.$ Find the magnetic moment of the bar magnet is (nearly)

• A. $5000{\text{Am}}^2$
• B. $3300{\text{Am}}^2$
• C. $490{\text{Am}}^2$
• D. $350{\text{Am}}^2$

Answer 1

The correct option is A $5000{\text{Am}}^2$

Time period of oscillation,

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

$\Rightarrow M=\frac{4{\pi }^2}{T^2}\times \frac{I}{B}$

$=\frac{4\times (3.14{)}^2\times 49\times {10}^{-2}}{(8.8{)}^2\times 0.5\times {10}^{-4}}$

$=5000{\text{Am}}^2$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.