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(B) The magnetic flux linked with the bigger loop due to the smaller loop is given by $\phi = M I$,where $M$ is the mutual inductance.Since the smalle

Vedclass · Accepted Answer

$9.1 	imes 10^{-11} \ Wb$

Vedclass · Answer

(B) The magnetic flux linked with the bigger loop due to the smaller loop is given by $\phi = M I$,where $M$ is the mutual inductance.Since the smaller loop is very small compared to the bigger loop,we can treat it as a magnetic dipole with magnetic moment $m = I_s A_s = I_s (\pi r^2)$.The magnetic field $B$ produced by the smaller loop at the center of the bigger loop is $B = \frac{\mu_0 m}{2(d^2 + R^2)^{3/2}}$,where $d = 0.15 \ m$,$R = 0.2 \ m$,and $r = 0.003 \ m$.However,it is easier to use the reciprocity theorem: $\phi_B = M I_s$,where $M$ is the same as the flux through the small loop due to unit current in the big loop.The magnetic field at the center of the small loop due to the big loop is $B_{big} = \frac{\mu_0 I_b R^2}{2(R^2 + d^2)^{3/2}}$.Flux $\phi = B_{big} 	imes A_{small} = \frac{\mu_0 I_b R^2}{2(R^2 + d^2)^{3/2}} 	imes \pi r^2$.Given $I_b = 2.0 \ A$,$R = 0.2 \ m$,$r = 0.003 \ m$,$d = 0.15 \ m$,and $\mu_0 = 4\pi 	imes 10^{-7} \ T \cdot m/A$.$\phi = \frac{(4\pi 	imes 10^{-7}) 	imes 2.0 	imes (0.2)^2}{2((0.2)^2 + (0.15)^2)^{3/2}} 	imes \pi (0.003)^2$.$\phi = \frac{4\pi^2 	imes 10^{-7} 	imes 2.0 	imes 0.04}{2(0.04 + 0.0225)^{3/2}} 	imes 9 	imes 10^{-6} = \frac{8\pi^2 	imes 10^{-7} 	imes 0.04}{2(0.0625)^{3/2}} 	imes 9 	imes 10^{-6}$.$(0.0625)^{3/2} = (0.25^2)^{3/2} = 0.25^3 = 0.015625$.$\phi = \frac{3.158 	imes 10^{-7}}{0.03125} 	imes 9 	imes 10^{-6} \approx 9.1 	imes 10^{-11} \ Wb$.

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