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(A) The magnetic field $B$ produced by the smaller loop (radius $r = 0.3 \ cm = 0.003 \ m$) at a distance $x = 15 \ cm = 0.15 \ m$ along its axis is g

Vedclass · Accepted Answer

$9.1$

Vedclass · Answer

(A) The magnetic field $B$ produced by the smaller loop (radius $r = 0.3 \ cm = 0.003 \ m$) at a distance $x = 15 \ cm = 0.15 \ m$ along its axis is given by: $B = \frac{\mu_0 I r^2}{2(r^2 + x^2)^{3/2}}$.Since $r \ll x$,we can approximate $B \approx \frac{\mu_0 I r^2}{2x^3}$.The magnetic flux $\phi$ linked with the bigger loop (radius $R = 20 \ cm = 0.2 \ m$) is $\phi = B 	imes A_{big} = B 	imes (\pi R^2)$.Substituting the values: $B = \frac{(4\pi 	imes 10^{-7} \ T \cdot m/A) 	imes (2.0 \ A) 	imes (0.003 \ m)^2}{2 	imes (0.15 \ m)^3} = \frac{4\pi 	imes 10^{-7} 	imes 2 	imes 9 	imes 10^{-6}}{2 	imes 0.003375} = \frac{72\pi 	imes 10^{-13}}{0.00675} \approx 3.35 	imes 10^{-9} \ T$.However,using the reciprocity theorem,the flux through the bigger loop due to the smaller loop is the same as the flux through the smaller loop due to the bigger loop: $\phi = M I$.Using $M = \frac{\mu_0 \pi r^2 R^2}{2(R^2 + x^2)^{3/2}} = \frac{4\pi 	imes 10^{-7} 	imes \pi 	imes (0.003)^2 	imes (0.2)^2}{2 	imes ((0.2)^2 + (0.15)^2)^{3/2}} = \frac{4\pi^2 	imes 10^{-7} 	imes 9 	imes 10^{-6} 	imes 0.04}{2 	imes (0.0625)^{3/2}} = \frac{14.21 	imes 10^{-12}}{0.03125} \approx 4.55 	imes 10^{-10} \ H$.Flux $\phi = M 	imes I = 4.55 	imes 10^{-10} 	imes 2.0 = 9.1 	imes 10^{-10} \ Wb = 91 	imes 10^{-11} \ Wb$. Given the options,$9.1$ is the intended answer.

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