The magnetic field due to a current-carrying | vedclass

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(B) Given: $R = 5 \ cm = 0.05 \ m$,$r = 12 \ cm = 0.12 \ m$,$B_{axis} = 250 \ \mu T$.The magnetic field due to a current-carrying circular loop at a p

Vedclass · Accepted Answer

$4394$

Vedclass · Answer

(B) Given: $R = 5 \ cm = 0.05 \ m$,$r = 12 \ cm = 0.12 \ m$,$B_{axis} = 250 \ \mu T$.The magnetic field due to a current-carrying circular loop at a point on its axis is given by:$B_{axis} = \frac{\mu_0 I R^2}{2(R^2 + r^2)^{3/2}}$The magnetic field at the centre of the loop is:$B_{centre} = \frac{\mu_0 I}{2R}$Dividing $B_{centre}$ by $B_{axis}$:$\frac{B_{centre}}{B_{axis}} = \frac{\mu_0 I}{2R} 	imes \frac{2(R^2 + r^2)^{3/2}}{\mu_0 I R^2} = \frac{(R^2 + r^2)^{3/2}}{R^3}$Substituting the values:$B_{centre} = B_{axis} 	imes \frac{(R^2 + r^2)^{3/2}}{R^3}$$B_{centre} = 250 \ \mu T 	imes \frac{(0.05^2 + 0.12^2)^{3/2}}{0.05^3}$$B_{centre} = 250 \ \mu T 	imes \frac{(0.0025 + 0.0144)^{3/2}}{0.000125}$$B_{centre} = 250 \ \mu T 	imes \frac{(0.0169)^{3/2}}{0.000125}$$B_{centre} = 250 \ \mu T 	imes \frac{(0.13)^3}{0.000125}$$B_{centre} = 250 \ \mu T 	imes \frac{0.002197}{0.000125} = 250 \ \mu T 	imes 17.576 = 4394 \ \mu T$.

The magnetic field due to a current-carrying circular loop of radius $5 \ cm$ at a point on the axis at a distance of $12 \ cm$ from the centre is $250 \ \mu T$. The magnetic field at the centre of the loop is (in $\mu T$)

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