Question

The magnitude of the magnetic induction at a point on the axis at a large distance $(r)$ from the centre of a circular coil of ${}^{\prime }{n}^{\prime }$ turns and area ${}^{\prime }{A}^{\prime }$ carrying current $(l)$ is given by

• A. $B_{axis}=\frac{{\mu }_0}{4\pi }\cdot \frac{nA}{l{r}^3}$
• B. $B_{axis}=\frac{{\mu }_0}{4\pi }\cdot \frac{2nlA}{r^3}$
• C. $B_{axis}=\frac{{\mu }_0}{4\pi }\cdot \frac{2nl}{A{r}^3}$
• D. $B_{axis}=\frac{{\mu }_0}{4\pi }\cdot \frac{nlA}{r^3}$

Answer 1

Answer: (B)

As we know that the magnetic field on the axis of a circular current carrying loop,
$B=\frac{{\mu }_{0}nI{a}^2}{2{\left(r^2+a^2\right)}^{3/2}}$...(i)
where, $I=$ current through the coil,
$a=$ radius of $a$ circular loop,
$r=$ distance of point from the centre along the axis
and $n=$ number of turns in the coil.
Area of the coil, $A=\pi a^2$
$\Rightarrow a^2=\frac{A}{\pi }$...(ii)
and it $r>>a$ then, ${\left(r^2+a^2\right)}^{3/2}\approx r^3$...(iii)
From Eqs. (i), (ii) and (iii), we get
$B=\left(\frac{{\mu }_{0}nI}{2r^3}\right)\frac{A}{\pi }\times \frac{2}{2}$
$\Rightarrow B=\frac{2{\mu }_{0}nIA}{4\pi r^3}$
So, option (b) is correct.