In an oscillation magnetometer, a bar magnet requires π/10 second to complete one oscillation. The moment of inertia of the magnet about its axis of rotation is 1.2 × 10^ -4 kg m^2 , and the earth's horizontal magnetic

Question

Quantrex Academy · Accepted Answer

Correct answer: D. 1600 A m^2

The time period of oscillation of a bar magnet in a magnetic field is given by:

T = 2\pi \sqrt{\dfrac{I}{MB_H}}

Squaring both sides, we get:

T^2 = \dfrac{4\pi^2 I}{MB_H}

Rearranging for the magnetic moment

M

:

M = \dfrac{4\pi^2 I}{T^2 B_H}

Substituting the given values

T = \dfrac{\pi}{10} \text{ s}

,

I = 1.2 \times 10^{-4} \text{ kg m}^2

, and

B_H = 30 \times 10^{-6} \text{ T}

:

M = \dfrac{4\pi^2 \times 1.2 \times 10^{-4}}{\left(\dfrac{\pi}{10}\right)^2 \times 30 \times 10^{-6}}

M = \dfrac{4\pi^2 \times 1.2 \times 10^{-4}}{\dfrac{\pi^2}{100} \times 30 \times 10^{-6}}

M = \dfrac{4 \times 1.2 \times 10^{-4} \times 100}{30 \times 10^{-6}}

M = \dfrac{4.8 \times 10^{-2}}{30 \times 10^{-6}}

M = 1600 \text{ A m}^2

In an oscillation magnetometer, a bar magnet requires π/10 second...

Options

Correct answer

Step-by-step solution

More from Permanent Magnets