Question

A thin bar magnet is cut into two equal parts as shown in figure the rotation of moment of inertia, magnetic moment of each part will become: (as compare to original magnet)
(A) ${\displaystyle \frac{1}{2}}$
(B) ${\displaystyle \frac{1}{4}}$
(C) ${\displaystyle \frac{1}{8}}$
(D) $1$

Answer 1

Hint :Here, the magnet given is acting like a rod so we have to use the concept of moment of inertia of rod along the axis of rotation from the middle of the bar magnet. Here, the extra property is the magnetic moment of the bar magnet. And we have to find the ratio between original and after cutting off the bar magnet from the middle as shown in the figure given.

Complete Step By Step Answer:
When the magnet is not cut from the between into two equal parts as shown in figure above.
Let $L$ be the length of the magnet and $M$ be the magnetic moment of the magnet.
Let pole strength be $S$ and $m$ be the mass of the magnet.
According to given data,
Initial moment of inertia $=m{L}^2$
Thus, initial magnetic moment $=S\times 2L$
Their ratios $={\displaystyle \frac{mL}{2S}}$ …….. $(1)$
After cutting the magnet pole strength remains the same.
Thus, final moment of inertia $={\displaystyle \frac{m}{2}}{\left({\displaystyle \frac{L}{2}}\right)}^2={\displaystyle \frac{m{L}^2}{8}}$
Magnetic moment $=S\times 2\left({\displaystyle \frac{L}{2}}\right)=S\times L$
Now, their ratios are given by $={\displaystyle \frac{mL}{8S}}$ ……. $(2)$
For the ratios of magnetic moment we have
Ratios of $(1)$ and $(2)$ $={\displaystyle \frac{{\displaystyle \frac{mL}{8S}}}{{\displaystyle \frac{mL}{2S}}}}={\displaystyle \frac{1}{4}}$
Thus magnetic moment of each part will become $1:4$
The correct answer is the option B.

Note :
We know how to calculate the moment of inertia of a rod that is why it is easier to calculate the moment of inertia of a bar magnet as it represents rod-like structure. Also magnetic moment is the moment calculated as a product of strength of pole and twice the length. Be careful while calculating.