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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
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The correct answer is:
$2 \sqrt{2}: 1$
Magnetic field at the centre of circular carrying conductor,
$$
B_{c}=\frac{\mu_{0} I}{2 r}
$$
Magnetic on the axis at a distance $x$ from the centre is given as
$$
B_{\text {axis }}=\frac{\mu_{0} I r^{2}}{2\left(r^{2}+x^{2}\right)^{3 / 2}}
$$
Here, $x=r$
$B_{\text {axis }}=\frac{\mu_{0} I r^{2}}{2\left(r^{2}+r^{2}\right)^{3 / 2}}$
$=\frac{\mu_{0} I r^{2}}{2\left(2 r^{2}\right)^{3 / 2}}=\frac{\mu_{0} I r^{2}}{2 \cdot 2 \sqrt{2} \cdot r^{3}}$
$B_{\text {axis }}=\frac{\mu_{0} I}{4 \sqrt{2} r}$
$B_{\text {axis }}=B_{a}=\frac{\mu_{0} I}{4 \sqrt{2} r}$
$\therefore \quad \frac{B_{c}}{B_{a}}=\frac{\mu_{0} I / 2 r}{\frac{\mu_{0} I}{4 \sqrt{2} r}}=\frac{4 \sqrt{2} r}{2 r}=2 \sqrt{2}$
$B_{c}: B_{a}=2 \sqrt{2}: 1$
$$
B_{c}=\frac{\mu_{0} I}{2 r}
$$
Magnetic on the axis at a distance $x$ from the centre is given as
$$
B_{\text {axis }}=\frac{\mu_{0} I r^{2}}{2\left(r^{2}+x^{2}\right)^{3 / 2}}
$$
Here, $x=r$
$B_{\text {axis }}=\frac{\mu_{0} I r^{2}}{2\left(r^{2}+r^{2}\right)^{3 / 2}}$
$=\frac{\mu_{0} I r^{2}}{2\left(2 r^{2}\right)^{3 / 2}}=\frac{\mu_{0} I r^{2}}{2 \cdot 2 \sqrt{2} \cdot r^{3}}$
$B_{\text {axis }}=\frac{\mu_{0} I}{4 \sqrt{2} r}$
$B_{\text {axis }}=B_{a}=\frac{\mu_{0} I}{4 \sqrt{2} r}$
$\therefore \quad \frac{B_{c}}{B_{a}}=\frac{\mu_{0} I / 2 r}{\frac{\mu_{0} I}{4 \sqrt{2} r}}=\frac{4 \sqrt{2} r}{2 r}=2 \sqrt{2}$
$B_{c}: B_{a}=2 \sqrt{2}: 1$
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