The magnetic field due to a current-carrying | vedclass

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Question

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

Vedclass · Accepted Answer

$250$

Vedclass · Answer

(A) The magnetic field on the axis of a circular loop is given by $B_{axis} = \frac{\mu_0 I r^2}{2(r^2 + x^2)^{3/2}}$.The magnetic field at the center of the loop is given by $B_{center} = \frac{\mu_0 I}{2r}$.Taking the ratio,we get $\frac{B_{center}}{B_{axis}} = \frac{(r^2 + x^2)^{3/2}}{r^3} = \left( 1 + \frac{x^2}{r^2} ight)^{3/2}$.Given $r = 3 \ cm$,$x = 4 \ cm$,and $B_{axis} = 54 \ \mu T$.Substituting the values: $\frac{B_{center}}{54} = \left( 1 + \left( \frac{4}{3} ight)^2 ight)^{3/2} = \left( 1 + \frac{16}{9} ight)^{3/2} = \left( \frac{25}{9} ight)^{3/2} = \frac{125}{27}$.Therefore,$B_{center} = 54 	imes \frac{125}{27} = 2 	imes 125 = 250 \ \mu T$.

The magnetic field due to a current-carrying circular loop of radius $3 \ cm$ at a point on the axis at a distance of $4 \ cm$ from the centre is $54 \ \mu T$. What will be its value at the centre of the loop? (in $\mu T$)

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