Question

The magnetic field due to current carrying circular coil loop of radius $6cm$ at a point on axis at a distance of $8cm$ from the centre is $54\mu T.$ What is the value at the centre of loop?

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

Answer 1

Answer: (D)

Given, radius of coil, $a=6cm$
$=6\times 10^{-2}m$
Distance, $r=8$
$=8\times 10^{-2}m$
Magnetic field on the axis of current $(I)$ carrying coil,
$B_1=\frac{{\left(\mu \right)}_{0}I{a}^2}{2{\left(a^2+r^2\right)}^{3/2}}$ ...(i)
Magnetic field on the centre of coil
$B_2=\frac{{\mu }_{0}I}{2a}$ ...(ii)
Dividing Equation (ii) by Equation (i), we get
$\frac{B_2}{B_1}=\frac{\frac{{\left(\mu \right)}_{0}I}{2a}}{\frac{{\left(\mu \right)}_{0}I{a}^2}{2{\left(a^2+r^2\right)}^{3/2}}}$
$=\frac{{\left(a^2+r^2\right)}^{3/2}}{a^3}$
$=\frac{{\left(\left[{\left(6\times {\left(10\right)}^{-2}\right)}^2+{\left(8\times {\left(10\right)}^{-2}\right)}^2\right]\right)}^{3/2}}{{\left(6\times {\left(10\right)}^{-2}\right)}^3}$
$\Rightarrow \frac{B_2}{B_1}=4.63$
$B_2=4.63\times B_1$
$=4.63\times 54\mu T$
$=250\mu T$