Question

The magnetic field due to a current carrying loop of radius $3cm$ at a point on its axis at a distance of $4cm$ from its centre is $54\mu T$. Then the value of the magnetic field at the centre of the loop is

• A. $250\mu T$
• B. $150\mu T$
• C. $75\mu T$
• D. $125\mu T$

Answer 1

Answer: (A)

$B_{\text{axis }}=54\times 10^{-6}=\frac{\mu NI{r}^2}{2{\left(r^2+x^2\right)}^{3/2}}$
Here, $r=3cm=3\times 10^{-2}m$
and $x=4cm=4\times 10^{-2}m$.
From above relation, we get
$\mu NI=\frac{54\times 10^{-2}\times 2\times {\left(25\times 10^{-4}\right)}^{3/2}}{{\left(3\times 10^{-2}\right)}^2}$
So, $B_{\text{centre }}=\frac{\mu NI}{2r}=250\mu T$