The magnetic field at the centre of a circula | vedclass

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(C) The magnetic field at the centre of a circular coil of radius $r$ carrying current $I$ is given by:$B_{c} = \frac{\mu_{0} I}{2r}$The magnetic fiel

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$2\sqrt{2} : 1$

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(C) The magnetic field at the centre of a circular coil of radius $r$ carrying current $I$ is given by:$B_{c} = \frac{\mu_{0} I}{2r}$The magnetic field on the axis of the circular coil at a distance $x$ from the centre is given by:$B_{a} = \frac{\mu_{0} I r^{2}}{2(r^{2} + x^{2})^{3/2}}$Given that the distance $x = r$,we substitute this into the expression for $B_{a}$:$B_{a} = \frac{\mu_{0} I r^{2}}{2(r^{2} + r^{2})^{3/2}} = \frac{\mu_{0} I r^{2}}{2(2r^{2})^{3/2}} = \frac{\mu_{0} I r^{2}}{2(2^{3/2} r^{3})} = \frac{\mu_{0} I}{2 \cdot 2\sqrt{2} \cdot r} = \frac{\mu_{0} I}{4\sqrt{2} r}$Now,calculating the ratio $B_{c} : B_{a}$:$\frac{B_{c}}{B_{a}} = \frac{\frac{\mu_{0} I}{2r}}{\frac{\mu_{0} I}{4\sqrt{2} r}} = \frac{4\sqrt{2} r}{2r} = 2\sqrt{2}$Therefore,the ratio $B_{c} : B_{a}$ is $2\sqrt{2} : 1$.

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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