Question

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

• A. $75 \, μT$
• B. $125 \, μT$
• C. $150 \, μT$
• D. $250 \, μT$

Answer 1

Answer: (D)

Given, radius of coil, $a=6 \, cm$
$=6\times 10^{- 2} \, m$
Distance, $r=8$
$=8\times 10^{- 2} \, m$
Magnetic field on the axis of current $\left(\right.I\left.\right)$ 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
$B_{2}=\frac{\mu _{0} I}{2 a}$ ...(ii)
Dividing Equation (ii) by Equation (i), we get
$\frac{B_{2}}{B_{1}}=\frac{\frac{\left(\mu \right)_{0} I}{2 a}}{\frac{\left(\mu \right)_{0} I a^{2}}{2 \left(a^{2} + r^{2}\right)^{3 / 2}} \, }$
$=\frac{\left(a^{2} + r^{2}\right)^{3 / 2}}{a^{3}}$
$=\frac{\left(\left[\left(6 \times \left(10\right)^{- 2}\right)^{2} + \left(8 \times \left(10\right)^{- 2}\right)^{2}\right]\right)^{3 / 2}}{\left(6 \times \left(10\right)^{- 2}\right)^{3}}$
$\Rightarrow \, \frac{B_{2}}{B_{1}}=4.63$
$B_{2}=4.63\times B_{1}$
$=4.63\times 54 \, μT$
$=250 \, μT$