A circular loop of radius 0.3 , cm lies paral | vedclass

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(A) The magnetic flux $\phi$ linked with the bigger loop due to the smaller loop is given by the formula for mutual induction between two loops where

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The magnetic flux $\phi$ linked with the bigger loop due to the smaller loop is given by the formula for mutual induction between two loops where the smaller loop acts as a magnetic dipole:$\phi = B \cdot A_{2} = \left( \frac{\mu_{0} I R_{1}^{2}}{2(R_{1}^{2} + x^{2})^{3/2}} ight) \cdot (\pi R_{2}^{2})$Here,$R_{1} = 0.3 \, cm = 0.3 	imes 10^{-2} \, m$ (radius of smaller loop),$R_{2} = 20 \, cm = 0.2 \, m$ (radius of bigger loop),$x = 15 \, cm = 0.15 \, m$ (distance between centers),$I = 20 \, A$ (current in smaller loop).Since $R_{1} \ll x$,we can approximate the field of the smaller loop as a dipole field,but using the exact formula for the flux through the larger loop due to the smaller one:$\phi = \frac{\mu_{0} I \pi R_{1}^{2} R_{2}^{2}}{2(R_{1}^{2} + x^{2})^{3/2}}$Substituting the values:$\phi = \frac{(4\pi 	imes 10^{-7}) 	imes 20 	imes \pi 	imes (0.3 	imes 10^{-2})^{2} 	imes (0.2)^{2}}{2((0.3 	imes 10^{-2})^{2} + (0.15)^{2})^{3/2}}$$\phi \approx \frac{4\pi^{2} 	imes 10^{-7} 	imes 20 	imes 9 	imes 10^{-6} 	imes 0.04}{2(0.15)^{3}}$$\phi \approx 9.1 	imes 10^{-11} \, Wb$.

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