Question

The magnetic field at the centre of a circular current carrying-conductor of radius $r$ is $B_{c}$ . The magnetic field on its axis at a distance r from the centre is $B_{a}$ . The value of $B_{c}:B_{a}$ will be :

• A. $1:\sqrt{2}$
• B. $1:2\sqrt{2}$
• C. $2\sqrt{2}:1$
• D. $\sqrt{2}:1$

Answer 1

Answer: (C)

Magnetic field at centre of current carrying coil,
$B_{c}=\frac{\left(\mu \right)_{0} I}{2 r}...\left(\right.i\left.\right)$
Magnetic field at axial point due to a currentcarrying coil at distance of $r$ ,
$d=r$
$B=\frac{\left(\mu \right)_{0} I r^{2}}{2 \left(r^{2} + d^{2}\right)^{3 / 2}}$
$\Rightarrow B_{a}=\frac{\left(\mu \right)_{0} I r^{2}}{2 \left(2 r^{2}\right)^{3 / 2}}....\left(\right.ii\left.\right)$
$Now,\frac{B_{c}}{B_{a}}=\frac{\left(\mu \right)_{0} I}{2 r}\times \frac{2 \left(2 r^{2}\right)^{3 / 2}}{\left(\mu \right)_{0} \left(Ir\right)^{2}}=2\sqrt{2}$
$B_{c}:B_{a}=2\sqrt{2}:1$