Question

A wire of fixed length is bent to form a coil of one turn. It is again bent to form a coil of three turns. If in both cases same amount of current is passed, then the ratio of the intensities of magnetic field produced at the centre of the coil of one turn and three turns respectively will be,

• A. $2:1$
• B. $9:1$
• C. $1:9$
• D. $1:3$

Answer 1

The correct option is C $1:9$

Magnetic field at the centre of a coil having $n$ and carrying a current $i$ is given by,

$B=\frac{n{\mu }_{0}i}{2r}$

For, $n=1,B=\frac{{\mu }_{0}i}{2r}$

If the same wire is again turned three times, then the radius of the new coil be $r^{\prime }$

$2\pi r=3\left(2\pi r^{\prime }\right)$

or $r^{\prime }=\frac{r}{3}$

Now, $n=3$, Hence the megnetic field at centre of the new coil will be

$B^{\prime }=\frac{n\left({\mu }_{0}i\right)}{2r^{\prime }}=\frac{3\left({\mu }_{0}i\right)}{2\left(\frac{r}{3}\right)}=\frac{9{\mu }_{0}i}{2r}$

$\Rightarrow B^{\prime }=9B$

$\frac{B}{B^{\prime }}=\frac{1}{9}$