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

Time period (T) of oscillation of a magnetic needle in a magnetic field is given by:

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

where: * I is the moment of inertia. * m is the magnetic moment. * B is the magnetic field.

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

Time for 10 oscillations = 10 s Time period (T) = $\frac{10}{10}$ = 1 s

1 = $2\pi \sqrt{\frac{10^{-6}/\pi^2}{(10^{-2})B}}$

1 = $2\sqrt{\frac{10^{-4}}{B}}$

B = 4 × 10^-4

T = 0.4 mT