Question

The magnetic moment and moment of inertia of a magnetic needle as shown are, respectively, \(1.0 \times 10^{-2} \, \text{A} \, \text{m}^2\) and \(\frac{10^{-6}}{\pi^2} \, \text{kg} \, \text{m}^2\). If it completes 10 oscillations in 10 s, the magnitude of the magnetic field is:

• A. 0.4 T
• B. 4 T
• C. 0.4 mT
• D. 4 mT

Answer 1

The Correct Option is C

The time period (T) of oscillation for a magnetic needle within a magnetic field is defined by the formula:

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

Where: * I represents the moment of inertia. * m denotes the magnetic moment. * B signifies the magnetic field strength.

Provided data: I = $\frac{10^{-6}}{\pi^2}$ kg m², m = 1.0 × 10^-2 A m²

Observation: 10 oscillations occurred in 10 seconds. Therefore, the time period (T) = $\frac{10}{10}$ = 1 second.

Substituting the given values into the formula yields: 1 = $2\pi \sqrt{\frac{10^{-6}/\pi^2}{(10^{-2})B}}$

This simplifies to: 1 = $2\sqrt{\frac{10^{-4}}{B}}$

Solving for B, we find: B = 4 × 10^-4

The magnetic field strength is calculated as B = 0.4 mT.