The magnetic field at the centre of a circula | vedclass

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Question

$B_{C}=\frac{\mu_{0} I}{2 r}, B_{a}=\frac{\mu_{0} I^{2}}{2\left(x^{2}+r^{2}\right)^{3 / 2}}$

$At \,\,x =\frac{r}{2}$

$B _{ a }=\frac{\mu_{0} Ir

Vedclass · Accepted Answer

$\left(\frac{2}{\sqrt{5}}\right)^{3} B$

Vedclass · Answer

$B_{C}=\frac{\mu_{0} I}{2 r}, B_{a}=\frac{\mu_{0} I^{2}}{2\left(x^{2}+r^{2}\right)^{3 / 2}}$

$At \,\,x =\frac{r}{2}$

$B _{ a }=\frac{\mu_{0} Ir ^{2}}{2\left(\frac{ r ^{2}}{4}+ r ^{2}\right)^{3 / 2}}$

$=\frac{\mu_{0} Ir ^{2}}{2\left(\frac{5}{4} r ^{2}\right)^{3 / 2}}=\frac{\mu_{0} I }{2 r }\left(\frac{4}{5}\right)^{3 / 2}$

$=\frac{\mu_{0} I }{2 r }\left(\frac{2}{\sqrt{5}}\right)^{3}$

The magnetic field at the centre of a circular coil of radius $I$, due to current I flowing through it, is $B$. The magnetic field at a point along the axis at a distance $\frac{r}{2}$ from the centre is

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