Magnetics Ques 138
- A hoop and a solid cylinder of same mass and radius are made of a permanent magnetic material with their magnetic moment parallel to their respective axes. But the magnetic moment of hoop is twice of solid cylinder. They are placed in a uniform magnetic field in such a manner that their magnetic moments make a small angle with the field. If the oscillation periods of hoop and cylinder are $T _h$ and $T _c$ respectively, then
(2019 Main, 10 Jan II)
(a) $T _h=0.5 T _c$
(b) $T _h=T _c$
(c) $T _h=2 T _c$
(d) $T _h=1.5 T _c$
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Answer:
Correct Answer: 138.(b)
Solution:
Formula:
Magnetic Moment Of A Current Carrying Loop:
- The time-period of oscillations made by a magnet of magnetic moment $M$, moment of inertia $I$, placed in a magnetic field is given by
$ T=2 \pi \sqrt{\frac{I}{M B}} $ $\quad$ …….(i)
For the hoop, let us assume its moment of inertia $I _h$ and magnetic moment $M _h$, then its time period will be
$ T _h=2 \pi \sqrt{\frac{I _h}{M _h B}} $ $\quad$ …….(ii)
Similarly, for solid cylinder, time period is,
$ T _c=2 \pi \sqrt{\frac{I _c}{M _c B}} $ $\quad$ …….(iii)
Dividing Eq. (ii) by Eq. (iii), we get
$ \frac{T _h}{T _c}=\sqrt{\frac{I _h M _c}{M _h I _c}} $ $\quad$ …….(iv)
Now, it is given that,
$ M _h=2 M _c $
and we know that, moment of inertia of hoop $I _h=m R^{2}$ and moment of inertia of solid
cylinder $I _c=\frac{1}{2} m R^{2}$
Substituting these values in Eq. (iv), we get
$ \begin{aligned} \frac{T _h}{T _c} & =\sqrt{\frac{m R^{2} \times M _c}{\frac{1}{2} m R^{2} \times 2 M _c}}=1 \\ \Rightarrow \quad T _h & =T _c \end{aligned} $
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