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The magnetic field at the centre of a circular current carrying conductor of radius $r$ is $\mathrm{B}_{\mathrm{c}}$. The magnetic field on its axis at a distance $r$ from the centre is $\mathrm{B}_{\mathrm{a}}$. The value of $\mathrm{B}_{\mathrm{c}}: \mathrm{B}_{\mathrm{a}}$ will be
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Verified Answer
The correct answer is:
$2 \sqrt{2}: 1$
Magnetic induction at the centre of the coil of radius is
$$
\mathrm{B}_{\mathrm{c}}=\frac{\mu_{0} \mathrm{nI}}{2 \mathrm{r}}
$$
Magnetic induction on the axial line of a circular coil at a distance from the centre is
$$
B_{a}=\frac{\mu_{0} n^{2} I}{2\left(r^{2}+x^{2}\right)^{3 / 2}}
$$
Given $x=r$
$$
B_{a}=\frac{\mu_{0} n r^{2} I}{2\left(2 r^{2}\right)^{3 / 2}}
$$
From Eqs. (i) and (ii), we get
$$
\frac{\mathrm{B}_{\mathrm{c}}}{\mathrm{B}_{\mathrm{a}}}=\frac{2 \sqrt{2}}{1}
$$
$$
\mathrm{B}_{\mathrm{c}}=\frac{\mu_{0} \mathrm{nI}}{2 \mathrm{r}}
$$
Magnetic induction on the axial line of a circular coil at a distance from the centre is
$$
B_{a}=\frac{\mu_{0} n^{2} I}{2\left(r^{2}+x^{2}\right)^{3 / 2}}
$$
Given $x=r$
$$
B_{a}=\frac{\mu_{0} n r^{2} I}{2\left(2 r^{2}\right)^{3 / 2}}
$$
From Eqs. (i) and (ii), we get
$$
\frac{\mathrm{B}_{\mathrm{c}}}{\mathrm{B}_{\mathrm{a}}}=\frac{2 \sqrt{2}}{1}
$$
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