Question

The magnetic field due to a current carrying loop of radius $3\, cm$ at a point on its axis at a distance of $4\, cm$ 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)

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