A bar magnet of mass 120 g, in the form of a rectangular parallelepiped, has dimensions l = 40 mm, b = 10 mm and h = 80 mm. With the dimension h vertical, the magnet performs angular oscillations in the plane of a magnetic field with period π s. If its magnetic moment is 3.4 A m²

Question

A bar magnet of mass 120 g, in the form of a rectangular parallelepiped, has dimensions l = 40 mm, b = 10 mm and h = 80 mm. With the dimension h vertical, the magnet performs angular oscillations in the plane of a magnetic field with period π s. If its magnetic moment is 3.4 A m², determine the influencing magnetic field.

Accepted Answer

> [!TLDR]
> T = π gives B = 4I/μ; I = 0.12(1600+100)/12 ×10⁻⁶ = 1.7×10⁻⁵; B = 4×1.7×10⁻⁵/3.4 = 2×10⁻⁵ T.

From the period of angular S.H.M.,

$$T = 2\pi\sqrt{\frac{I}{\mu B}} \Rightarrow \pi = 2\pi\sqrt{\frac{I}{\mu B}} \Rightarrow B = \frac{4I}{\mu}$$

The moment of inertia of the parallelepiped about the vertical axis (l and b horizontal) is

$$I = M\frac{l^2 + b^2}{12} = 0.12	imes\frac{(40^2 + 10^2)	imes10^{-6}}{12} = 1.7	imes10^{-5}\ \mathrm{kg\,m^2}$$

$$	herefore B = \frac{4	imes 1.7	imes10^{-5}}{3.4} = 2	imes10^{-5}\ \mathrm{Wb\,m^{-2}}\ (\mathrm{T})$$

> [!ANSWER]
> I = 1.7×10⁻⁵ kg m², B = 2×10⁻⁵ T (Wb m⁻²).

🎓Unlock everything free for 14 days

Unlock everything free for 14 days