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The magnetic field at the center of a current carrying loop of radius \( 0.1 \mathrm{~m} \) is \( 5 \sqrt{5} \) times that at
a point along its axis. The distance of this point from the centre of the loop is
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a point along its axis. The distance of this point from the centre of the loop is
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The correct answer is:
\( 0.2 \mathrm{~m} \)
Magnetic field at the centre of current carrying loop is given $\operatorname{as} B_{C}=\frac{\mu_{0}}{2} \frac{I}{r}$
Magnetic field at the axis is given as $B_{A}=\frac{\mu_{0}}{2} \frac{I r^{2}}{\left(\chi^{2}+r^{2}\right)^{3 / 2}}$
Given, $B_{C}=5 \sqrt{5} B_{A}$
$\Rightarrow \frac{\mu_{0}}{2} \frac{I}{r}=5 \sqrt{5} \frac{\mu_{0}}{2} \frac{I r^{2}}{\left(\chi^{2}+r^{2}\right)^{3 / 2}}$
$\Rightarrow\left(x^{2}+r^{2}\right)^{3 / 2}=r^{3} 5 \sqrt{5}$
$\Rightarrow\left(x^{2}+r^{2}\right)^{3 / 2}=r^{3} \times 5^{3 / 2}$
$\Rightarrow \chi^{2}+r^{2}=r^{2} \times 5$
$\Rightarrow \chi^{2}=5 r^{2}-r^{2}$
$\Rightarrow \chi^{2}=4 r^{2}$ or $x=2 r$
Given, $r=0.1 \mathrm{~m}$
$\Rightarrow \chi=0.2 m$
Magnetic field at the axis is given as $B_{A}=\frac{\mu_{0}}{2} \frac{I r^{2}}{\left(\chi^{2}+r^{2}\right)^{3 / 2}}$
Given, $B_{C}=5 \sqrt{5} B_{A}$
$\Rightarrow \frac{\mu_{0}}{2} \frac{I}{r}=5 \sqrt{5} \frac{\mu_{0}}{2} \frac{I r^{2}}{\left(\chi^{2}+r^{2}\right)^{3 / 2}}$
$\Rightarrow\left(x^{2}+r^{2}\right)^{3 / 2}=r^{3} 5 \sqrt{5}$
$\Rightarrow\left(x^{2}+r^{2}\right)^{3 / 2}=r^{3} \times 5^{3 / 2}$
$\Rightarrow \chi^{2}+r^{2}=r^{2} \times 5$
$\Rightarrow \chi^{2}=5 r^{2}-r^{2}$
$\Rightarrow \chi^{2}=4 r^{2}$ or $x=2 r$
Given, $r=0.1 \mathrm{~m}$
$\Rightarrow \chi=0.2 m$
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