Question

A current passing through a circular coil of two turns produces a magnetic field B as its centre. The coil is then rewound so as to have four turns and the same current is passed through it. The magnetic field at its centre now is:

• A. $2B$
• B. $\frac{B}{2}$
• C. $\frac{B}{4}$
• D. $4B$

Answer 1

Answer: (D)

$4B$

Let the radius of the initial circular coil be $r$

Thus length of the coil $L=2\left(2\pi r\right)$

Now the coil is rewound so as to have $4$ turns such that the length of the coil is same.

$\therefore$ $L=4\left(2\pi r^{\prime }\right)=2\left(2\pi r\right)$ $⟹r^{\prime }=\frac{r}{2}$

Magnetic field at the centre of the coil $B_c=\frac{{\mu }_o}{4\pi }\frac{i}{r}\left(2\pi n\right)$ where $n$ is the number of turns of the coil.

For 1st case : $n=2$ $⟹B=\frac{{\mu }_o}{4\pi }\frac{i}{r}(2\pi \times 2)=\frac{{\mu }_{o}i}{r}$ .........(1)

For 2nd case : $n^{\prime }=4$ $r^{\prime }=\frac{r}{2}$

$⟹B^{\prime }=\frac{{\mu }_o}{4\pi }\frac{i}{\frac{r}{2}}(2\pi \times 4)=4\frac{{\mu }_{o}i}{r}$

From (1), we get $B^{\prime }=4B$