Question

The magnetic field due to a current carrying circular loop of radius 3 cm at a point on its axis at a distance of 4 cm from the centre is 54 $\mu$T. The magnetic field (in $\mu$T) at the centre of the loop will be

• A. 250
• B. 150
• C. 125
• D. 72

Answer 1

The correct option is A

250

Given x = 4 cm = 0.04 m and r = 3 cm = 0.03 m.
The magnetic field at a point on the axis of the loop is given by
$B=\frac{{\mu }_{0}I{r}^2}{2(r^2+x^2{)}^{\frac{3}{2}}}$ (1)
Magnetic field at the centre of the coil is given by
$B_0=\frac{{\mu }_{0}I}{2r}$ (2)
Dividing (1) by (2), we get
$\frac{B_0}{B}=\frac{r^3}{(r^2+x^2{)}^{\frac{3}{2}}}$
Substituting the values of r and x, we get
$\frac{B_0}{B}=\frac{125}{27}or{B}_0=\frac{125}{27}B=\frac{125}{27}\times 54\mu T=250\mu T$
Hence the correct choice is (a).