Question

In a uniform magnetic field, the magnetic needle has a magnetic moment $9.85 \times 10^{-2}$ A/m$^2$ and moment of inertia $5 \times 10^{-6}$ kg m$^2$. If it performs 10 complete oscillations in 5 seconds then the magnitude of the magnetic field is _________ mT. [Take $\pi^2$ as 9.85]

Hint:

This problem is a direct application of the formula for the time period of a magnetic compass in a B-field. Memorize the formula $T = 2\pi\sqrt{I/mB}$ and be careful with algebraic rearrangement and unit conversions (Tesla to milliTesla).

Answer 1

Correct Answer: 8

First, we calculate the time period (T) of the oscillation.
The needle performs 10 oscillations in 5 seconds.
$T = \frac{\text{Total time}}{\text{Number of oscillations}} = \frac{5 \text{ s}}{10} = 0.5$ s.
The formula for the time period of a magnetic dipole oscillating in a magnetic field B is:
$T = 2\pi\sqrt{\frac{I}{mB}}$
where I is the moment of inertia and m is the magnetic moment.
We need to find B. Squaring both sides and rearranging the formula:
$T^2 = 4\pi^2 \frac{I}{mB} \implies B = \frac{4\pi^2 I}{mT^2}$
We are given the values:
$I = 5 \times 10^{-6}$ kg m$^2$
$m = 9.85 \times 10^{-2}$ A/m$^2$
$T = 0.5$ s
$\pi^2 = 9.85$
Substitute these values into the equation for B:
$B = \frac{4 \times (9.85) \times (5 \times 10^{-6})}{(9.85 \times 10^{-2}) \times (0.5)^2}$
Cancel out the $9.85$ term:
$B = \frac{4 \times 5 \times 10^{-6}}{10^{-2} \times 0.25} = \frac{20 \times 10^{-6}}{0.25 \times 10^{-2}}$
$B = \frac{20}{0.25} \times 10^{-4} = 80 \times 10^{-4} \text{ T} = 8 \times 10^{-3}$ T.
The question asks for the answer in milliTesla (mT).
$B = 8 \times 10^{-3} \text{ T} = 8$ mT.