Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

A coil carrying current 1 has radius $r$ and number of units $n$. It is rewound so that the radius of the new coil is $\dfrac{r}{4}$ and it carries current $1$. The ratio of magnetic moment of new coil to that of original coil is
A. $1$
B. $\dfrac{1}{2}$
C. $\dfrac{1}{4}$
D. $\dfrac{1}{8}$

seo-qna
Last updated date: 22nd May 2024
Total views: 332.7k
Views today: 4.32k
Answer
VerifiedVerified
332.7k+ views
Hint:Coil carrying current radius and number of turns is given and new coil turns radius is given, measure the ratio of magnetic moment.he magnetic moment is a vector quantity. The objects have a tendency to place themselves in such a way that the magnetic moment vector becomes parallel to the magnetic field lines. The direction of the magnetic moment points from the south to the north pole of a magnet.

Complete step by step answer:
Initially radius of the coil is $r$ and number of turns of the coil is $n$ respectively,
The initial length of the wire is $L=n\left( {2\pi r} \right)$
Area of the coil is, A=$\pi {r^2}$
From the initial magnetic moment, $\mu = nlA$
Already $A$ value is the area of the coil, then
$\mu = nl\pi {r^2} \to \left( 1 \right)$
Now the same wire is wound to the coil of radius, ${r_1} = \dfrac{r}{4}$
Assuming the number of turns of the coil is, ${n_1}$
The length of the wire is, ${L_1} = {n_1}\left( {2\pi {r_1}} \right)$
Substituting the value of ${r_1}$, then
${L_1} = \dfrac{{2\pi {n_1}r}}{4}$
Suppose ${L_1} = L$ then,
$2\pi rn = \dfrac{{2\pi r{n_1}}}{4} \\
\Rightarrow {n_1} = 4n \\ $
Now area of new coil is
${A_1} = \pi {\left( {{r_1}} \right)^2} \\
\Rightarrow {A_1} = \dfrac{{\pi {r^2}}}{{16}} \\ $
Let us calculate the magnetic moment of the new coil then we get,
${\mu _1} = {n_1}l{A_1}$
We already calculated all the values in left hand side, by substituting those values in the above equation we get,
${\mu _1} = 4nl \times \dfrac{{\pi {r^2}}}{{16}} \\
\Rightarrow {\mu _1} = \dfrac{{nl\pi {r^2}}}{4} \\ $
Therefore,
${\mu _1} = \dfrac{\mu }{4} \\
\therefore \dfrac{{{\mu _1}}}{\mu } = \dfrac{1}{4} \\ $
Hence the ratio of magnetic moment of the new coil to that of the original coil is, $\dfrac{1}{4}$ and the correct option is C.

Note: Current carrying coil and its radius and number of turns are different from the number of turns of the coil and radius of the coil that carries current of the new coil. Simple defined as, the ratio of magnetic moment of new coil to the magnetic moment of that original coil, which carries different current in both the coils.
Recently Updated Pages