Question

A coil of radius R carries a current I. Another concentric coil of radius r ( r < < R) carries current $\frac{I}{2}$. Initially planes of the two coils are mutually perpendicular and both the coils are free to rotate with M and m respectively (m < M). During the subsequent motion, let $K_{1}and{K}_2$ be the maximum kinetic energies of the two coils respectively and let $U$ be the magnitude of maximum potential energy of magnetic interaction of the system of the coils. Choose the correct options :

• A. $\frac{K_1}{K_2}=\frac{M}{m}{\left(\frac{R}{r}\right)}^2$
• B. $K_1=\frac{Um{r}^2}{m{r}^2+M{R}^2},K_2=\frac{UM{R}^2}{m{r}^2+M{R}^2}$
• C. $U=\frac{{\mu }_0\pi I^2.r^2}{4R}$
• D. $K_2>>K_1$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $K_1=\frac{Um{r}^2}{m{r}^2+M{R}^2},K_2=\frac{UM{R}^2}{m{r}^2+M{R}^2}$
C $U=\frac{{\mu }_0\pi I^2.r^2}{4R}$
D $K_2>>K_1$

Magnetic field at the centre due to circular ring of radius $R$ carrying current $I$

$B=\frac{{\mu }_0{I}_2}{2R}$. . . . .(1)

The dipole moment of circle of radius $r$ carrying current $\frac{I}{2}$ is,

$\mu =\pi r^2.\frac{I}{2}=\frac{\pi r^{2}I}{2}$. . . . .(2)

Initial potential energy,

$U_i=-\mu Bcos\theta =\mu Bcos{90}^0=0$

Final potential energy in magnetic field,

$U_f=-\mu B=\frac{{\mu }_0\pi r^2{I}^2}{4R}$ (by putting the value of $B$ and $\mu$ )

Change in potential energy in magnetic field,

$\Delta U=|U_f-U_i|=\mu B$

$\Delta U=\frac{{\mu }_0\pi r^2{I}^2}{4R}=U$ which convert into kinetic energy.. . . . .(3)

In rotational mechanics,

by conservation of angular momentum,

$i_1{\omega }_1=i_2{\omega }_2$, where here, $i_1$ and $i_2$ are moment of inertia of radius of circle $R$ and $r$ respectively

${\omega }_1=\frac{i_2{\omega }_2}{i_2}$

Conservation of energy,

$\frac{1}{2}{i}_1{\omega }_1^2+\frac{1}{2}{i}_2{\omega }_2^2=\Delta U=U$. . . . .(4)

put the value of ${\omega }_1$ in the above equation, we get,

$\frac{1}{2}{i}_2{\omega }_2^2=\frac{i_1\Delta U}{i_1+i_2}=K_2$(Kinetic energy of small ring). . . . . .(5)

moment of inertia of ring of radius $R$ is

$i_1=\frac{1}{2}M{R}^2$

moment of inertia of ring of radius $r$ is,

$i_2=\frac{1}{2}m{r}^2$

Put the value of $i_1$,$i_2$ and $\Delta U$ in equation (4), we get

$K_2=\frac{UM{R}^2}{M{R}^2+m{r}^2}$

Put $K_2$ in the equation (4), we get

$K_1=\frac{1}{2}{i}_1{\omega }_1=\frac{Um{r}^2}{M{R}^2+m{r}^2}$

Given, $M>m$ and $R>r$

Kinetic energy of ring is depended on mass and radius,

So, $K_2>>K_1$

Thus, the correct answer is B, C and D.