Question

A magnet in the form of a cylindrical rod has a length of 5.51 cm and a diameter of 0.865 cm. It has a uniform magnetization of 5.17 x 103 A/m. The magnetic dipole moment?

Answer 1

Answer:

Magnetic dipole moment will be$15.51\times 10^{-3}J/T$

Explanation:

We have given length of the cylinder , that is h = 5.51 cm = 0.051 m

And diameter of the cylinder d = 0.865 cm

So radius $r=\frac{d}{2}=\frac{0.865}{2}=0.4325cm=0.4325\times 10^{-2}m$

So volume of cylinder $V=\pi r^2h=3.14\times (0.4325\times 10^{-2})^2\times 0.051=3\times 10^{-6}m^3$

It is given there is uniform magnetization of $M=5.17\times 10^3A/m$

We have to fond the dipole moment

Dipole moment is equal to $\mu =MV$, here M is magnetization and V is volume

So $\mu =MV=5.17\times 10^3\times 3\times 10^{-6}=15.51\times 10^{-3}J/T$