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 find the value of \(x\) for the magnetic moment \(M = x \times 10^{-5} \, \text{Am}^2\), we start with the formula for the time period \(T\) of oscillation of a magnetic needle:

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

where \(I\) is the moment of inertia, \(B\) is the magnetic field, and \(M\) is the magnetic moment.

The needle performs \(20\) oscillations in \(5\) seconds, so the time period for one oscillation is:

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

Substitute the given 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 formula:

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

Square both sides to eliminate the square root:

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

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

Rearrange and solve for \(M\):

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

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

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

Answer 2

Calculating the Magnetic Moment (M)

Step 1: Recall the Formula for the Time Period of a Magnetic Needle

The time period \( T \) is given by:

T = \(2\pi \sqrt{\frac{I}{MB}}\)

Where \(I\) is the moment of inertia, \(M\) is the magnetic moment, and \(B\) is the magnetic field.

Step 2: Calculate the Time Period for One Oscillation

Given 20 oscillations in 5 seconds:

T = 5 / 20 = 0.25 s

Step 3: Rearrange the Formula to Find M

Rearranging the formula to solve for \( M \):

\(M = \frac{4\pi^2 I}{T^2 B}\)

Step 4: Substitute the Values

Substituting the values:

\(M = \frac{4\pi^2 (9.8 \times 10^{-6})}{(0.25)^2 (0.049)}\)

\(M = \frac{4\pi^2 (9.8 \times 10^{-6})}{0.0030625}\)

\(M = 128 \pi^2 \times 10^{-5} \, \text{A m}^2\)

Step 5: Conclude

The value of \( x \) is: \(1280 \pi^2\)