Question

A circular coil connected to a cell of e.m.f $E$ produced a magnetic field. The coil is unwound, stretched to double its length, rewound into a coil of $\frac{1}{3}$ of the original radius and connected to a cell of e.m.f $E_1$ to produce the same field at the centre. Then $E_1$ is :

• A. $\frac{E}{2}$
• B. $\frac{2E}{3}$
• C. $\frac{9E}{4}$
• D. $\frac{E}{6}$

Answer 1

Answer: (B)

$\frac{2E}{3}$

Let initial length is $L$ and Area of cross-section of wire is $A$. Consider $L$ and $A$ are the length and area after stretching.

Then, volumeof wire must remain same, i.e. $A^{\prime }L=A{L}^{\prime }$

$A^{\prime }L=2AL$

$A^{\prime }=\frac{A}{2}$

Therefore,
$\frac{R_i}{R_f}=\frac{LA}{A.L}=\frac{1}{4}$

Where $R_i$ and $R_f$ are initial and final resistance

Or, $R_f=4R_i$

Assuming the number of turns are also doubled by stretching the length and circumference of coil is sufficient to accommodate even after the coil radius is reduced to $\frac{1}{3}$ of original radius

Now,
Magnetic field at the center of the coil is same.

Therefore,

$\frac{I}{r}=6\frac{I_f}{r}$ Where $I$ and $I_f$ are initial and final current

Or,$I=6I_f$

$\frac{E}{E_1}=\frac{I{R}_i}{I_f{R}_f}=\frac{3}{2}$

Or, $E_1=\frac{2E}{3}$