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Question: A coil carrying current 1 has radius \(r\) and number of units \(n\). It is rewound so that the radi...

A coil carrying current 1 has radius rr and number of units nn. It is rewound so that the radius of the new coil is r4\dfrac{r}{4} and it carries current 11. The ratio of magnetic moment of new coil to that of original coil is
A. 11
B. 12\dfrac{1}{2}
C. 14\dfrac{1}{4}
D. 18\dfrac{1}{8}

Explanation

Solution

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 rr and number of turns of the coil is nn respectively,
The initial length of the wire is L=n(2πr)L=n\left( {2\pi r} \right)
Area of the coil is, A=πr2\pi {r^2}
From the initial magnetic moment, μ=nlA\mu = nlA
Already AA value is the area of the coil, then
μ=nlπr2(1)\mu = nl\pi {r^2} \to \left( 1 \right)
Now the same wire is wound to the coil of radius, r1=r4{r_1} = \dfrac{r}{4}
Assuming the number of turns of the coil is, n1{n_1}
The length of the wire is, L1=n1(2πr1){L_1} = {n_1}\left( {2\pi {r_1}} \right)
Substituting the value of r1{r_1}, then
L1=2πn1r4{L_1} = \dfrac{{2\pi {n_1}r}}{4}
Suppose L1=L{L_1} = L then,
2πrn=2πrn14 n1=4n 2\pi rn = \dfrac{{2\pi r{n_1}}}{4} \\\ \Rightarrow {n_1} = 4n \\\
Now area of new coil is
A1=π(r1)2 A1=πr216 {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,
μ1=n1lA1{\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,
μ1=4nl×πr216 μ1=nlπr24 {\mu _1} = 4nl \times \dfrac{{\pi {r^2}}}{{16}} \\\ \Rightarrow {\mu _1} = \dfrac{{nl\pi {r^2}}}{4} \\\
Therefore,
μ1=μ4 μ1μ=14 {\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, 14\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.