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The ratio of the magnetic fields at the centre of a circular coil carrying current to that at a point whose distance is half of the radius of the coil is
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The correct answer is:
$5 \sqrt{5}: 8$
The magnetic field due to a circular coil carrying current at its axis is
$$
B_{x}=\frac{\mu_{0}}{4 \pi} \cdot \frac{2 \pi N i r^{2}}{\left(x^{2}+r^{2}\right)^{3 / 2}} \quad \text{...(i)}
$$
At centre of coil, $B_{c}=\frac{\mu_{0}}{4 \pi} \frac{2 \pi N i}{r} \quad \text{...(ii)}$
From Eqs. (i) and (ii), we get
$\frac{B_{c}}{D_{x}}=\frac{\left(x^{2}+r^{2}\right)^{3 / 2}}{r \times r^{2}}$
At, $\quad x=\frac{r}{2}, \frac{B_{c}}{B_{x}}=\frac{\left(\frac{r^{2}}{4}+r^{2}\right)^{3 / 2}}{r^{3}}=\frac{\left(\frac{5}{4}\right)^{3 / 2} r^{3}}{r^{3}}=\frac{5 \sqrt{5}}{8}$
or $5 \sqrt{5}: 8$
$$
B_{x}=\frac{\mu_{0}}{4 \pi} \cdot \frac{2 \pi N i r^{2}}{\left(x^{2}+r^{2}\right)^{3 / 2}} \quad \text{...(i)}
$$
At centre of coil, $B_{c}=\frac{\mu_{0}}{4 \pi} \frac{2 \pi N i}{r} \quad \text{...(ii)}$
From Eqs. (i) and (ii), we get
$\frac{B_{c}}{D_{x}}=\frac{\left(x^{2}+r^{2}\right)^{3 / 2}}{r \times r^{2}}$
At, $\quad x=\frac{r}{2}, \frac{B_{c}}{B_{x}}=\frac{\left(\frac{r^{2}}{4}+r^{2}\right)^{3 / 2}}{r^{3}}=\frac{\left(\frac{5}{4}\right)^{3 / 2} r^{3}}{r^{3}}=\frac{5 \sqrt{5}}{8}$
or $5 \sqrt{5}: 8$
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