Question

In a uniform magnetic field of \(0.049 T\), a magnetic needle performs \(20\) complete oscillations in \(5\) seconds as shown. The moment of inertia of the needle is \(9.8 \times 10 kg m^2\). If the magnitude of magnetic moment of the needle is \(x \times 10^{-5} Am^2\); then the value of '\(x\)' is
 

• A. \(5\pi^2\)
• B. \(128\pi^2\)
• C. \(50\pi^2\)
• D. \(1280\pi^2\)

Answer 1

The Correct Option is D

To determine the value of \(x\) in the magnetic moment \(M = x \times 10^{-5} \, \text{Am}^2\), we employ the formula for the period of oscillation \(T\) of a magnetic needle:

\[T=2\pi\sqrt{\frac{I}{MB}}\]

Here, \(I\) represents the moment of inertia, \(B\) is the magnetic field strength, and \(M\) denotes the magnetic moment.

The needle completes \(20\) oscillations in \(5\) seconds. Therefore, the time period for a single oscillation is calculated as:

\[T=\frac{5 \, \text{s}}{20} = \frac{1}{4} \, \text{s}\]

Substitute the provided values: \(I=9.8 \times 10 \, \text{kg m}^2\), \(T=\frac{1}{4} \, \text{s}\), and \(B=0.049 \, \text{T}\) into the time period equation:

\[\frac{1}{4}=2\pi\sqrt{\frac{9.8 \times 10}{M\times 0.049}}\]

Square both sides of the equation to remove the radical:

\[\left(\frac{1}{4}\right)^2=4\pi^2\frac{9.8 \times 10}{M\times 0.049}\]

\[\frac{1}{16}=4\pi^2\frac{98}{M\times 0.049}\]

Rearrange the equation to solve for \(M\):

\[M=\frac{4\pi^2\times 98\times 16}{0.049}\]

\[M=1280\pi^2 \times 10^{-5}\]

Consequently, the value of \(x\) is \(\boxed{1280\pi^2}\).