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. $\dfrac{1}{8}$
Last updated date: 29th Mar 2023
•
Total views: 207k
•
Views today: 1.84k
Answer
207k+ 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.
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
Most eubacterial antibiotics are obtained from A Rhizobium class 12 biology NEET_UG

Salamin bioinsecticides have been extracted from A class 12 biology NEET_UG

Which of the following statements regarding Baculoviruses class 12 biology NEET_UG

Sewage or municipal sewer pipes should not be directly class 12 biology NEET_UG

Sewage purification is performed by A Microbes B Fertilisers class 12 biology NEET_UG

Enzyme immobilisation is Aconversion of an active enzyme class 12 biology NEET_UG

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE
