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 $B_{	ext{axis}} = \frac{\mu_0 I r^2}{2(r^2 + x^2)^{3/2}}$.Given $r = 3 	ext{ cm}$

Vedclass · Accepted Answer

$250$

Vedclass · Answer

(A) The magnetic field on the axis of a circular loop is given by $B_{	ext{axis}} = \frac{\mu_0 I r^2}{2(r^2 + x^2)^{3/2}}$.Given $r = 3 	ext{ cm}$ and $x = 4 	ext{ cm}$,the distance from the centre is $d = \sqrt{r^2 + x^2} = \sqrt{3^2 + 4^2} = 5 	ext{ cm}$.Substituting the values: $54 \mu	ext{T} = \frac{\mu_0 I r^2}{2(5 	ext{ cm})^3} = \frac{\mu_0 I (3 	ext{ cm})^2}{2(125 	ext{ cm}^3)}$.Thus,$\frac{\mu_0 I}{2} = \frac{54 	imes 125}{9} = 6 	imes 125 = 750 \mu	ext{T} \cdot 	ext{cm}$.The magnetic field at the centre is $B_{	ext{centre}} = \frac{\mu_0 I}{2r} = \frac{750 \mu	ext{T} \cdot 	ext{cm}}{3 	ext{ cm}} = 250 \mu	ext{T}$.

The magnetic field due to a current-carrying loop of radius $3 \text{ cm}$ at a point on its axis at a distance of $4 \text{ cm}$ from its centre is $54 \mu\text{T}$. Then,the value of the magnetic field at the centre of the loop is: (in $\mu\text{T}$)

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