Question

The radius of a coil of wire with $N$ turns is $0.22\text{m}$ and $3.5\text{A}$ current flows clockwise as shown in figure. A long straight wire is carrying a current $54\text{A}$ towards the left as shown. The shortest distance between the straight wire and the coil is $0.05\text{m}$. The magnetic field at the centre of coil is zero. The number of turns $N$ in the coil is:

• A. $4$
• B. $6$
• C. $7$
• D. $8$

Answer 1

The correct option is A $4$

The distance of centre of coil from the straight long wire is
$d=(r+0.05)\text{m}$


The direction of field at centre due to circular coil is in perpendicularly inward $⨂$ direction & due to straight wire in perpendicularly outward $⨀$ direction.
$B_{centre}=0$

$\Rightarrow B_1-B_2=0$

or, $B_1=B_2...........\left(1\right)$

Here $B_1$ is magnetic field due to coil at centre and $B_2$ is the magnetic field due to straight wire at the centre.

$B_1=\frac{{\mu }_{0}N{i}_1}{2r}=\frac{\left({\mu }_{0}N\right)\left(3.5\right)}{2\left(0.22\right)}$

Magnetic field due to straight wire $B_2$ is,

$B_2=\frac{{\mu }_0{i}_2}{2\pi d}=\frac{{\mu }_0\left(54\right)}{2\pi (r+0.05)}$

From Eq $\left(1\right)$ we have,

$\frac{{\mu }_0\times \left(54\right)}{(2\times \frac{22}{7})(r+0.05)}=\frac{\left({\mu }_{0}N\right)\left(350\right)}{2\times 22}$

or, $\frac{{\mu }_0\left(54\right)\times 7}{44\times (r+0.05)}=\frac{{\mu }_0\times 350N}{44}$

or, $\frac{54}{r+0.05}=\frac{50N}{1}$

or, $\frac{54}{0.27}=50N$

or, $\frac{54\times 100}{27}=50N$

$\therefore N=\frac{200}{50}=4$