The magnetic field due to a current-carrying | vedclass

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(B) The magnetic field on the axis of a circular loop is given by: $B_{	ext{axis}} = \frac{\mu_0 I a^2}{2(a^2 + x^2)^{3/2}}$The magnetic field at the

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$1000$

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(B) The magnetic field on the axis of a circular loop is given by: $B_{	ext{axis}} = \frac{\mu_0 I a^2}{2(a^2 + x^2)^{3/2}}$The magnetic field at the centre of the loop is given by: $B_{	ext{centre}} = \frac{\mu_0 I}{2a}$Taking the ratio of the two fields:$\frac{B_{	ext{centre}}}{B_{	ext{axis}}} = \frac{\mu_0 I}{2a} 	imes \frac{2(a^2 + x^2)^{3/2}}{\mu_0 I a^2} = \frac{(a^2 + x^2)^{3/2}}{a^3}$Given $a = 6 \ cm$ and $x = 8 \ cm$:$\frac{B_{	ext{centre}}}{B_{	ext{axis}}} = \frac{(6^2 + 8^2)^{3/2}}{6^3} = \frac{(36 + 64)^{3/2}}{216} = \frac{(100)^{3/2}}{216} = \frac{1000}{216}$Given $B_{	ext{axis}} = 216 \ \mu T$:$B_{	ext{centre}} = 216 	imes \frac{1000}{216} = 1000 \ \mu T$.

The magnetic field due to a current-carrying circular loop of radius $6 \ cm$ at a point on the axis at a distance of $8 \ cm$ from the centre is $216 \ \mu T$. Then the magnetic field at the centre of the ring is $\dots \ \mu T$.

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