Two concentric circular coils,C_{1} and C_{2} | vedclass

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Question

(B) The magnetic field $B$ at the center of coil $C_{2}$ due to current $I$ is given by $B = \frac{\mu_{0} N_{2} I}{2 R_{2}}$.The magnetic flux $\phi$

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) The magnetic field $B$ at the center of coil $C_{2}$ due to current $I$ is given by $B = \frac{\mu_{0} N_{2} I}{2 R_{2}}$.The magnetic flux $\phi$ linked with coil $C_{1}$ is $\phi = B \cdot A_{1} \cdot N_{1} = \left( \frac{\mu_{0} N_{2} I}{2 R_{2}} ight) (\pi r_{1}^{2}) N_{1}$.Substituting the values: $N_{1} = 500$,$r_{1} = 0.01\; m$,$N_{2} = 200$,$R_{2} = 0.2\; m$,$\mu_{0} = 4\pi 	imes 10^{-7}\; T\cdot m/A$.$\phi = \frac{(4\pi 	imes 10^{-7}) 	imes 200 	imes (5t^{2} - 2t + 3)}{2 	imes 0.2} 	imes \pi 	imes (0.01)^{2} 	imes 500$.$\phi = \frac{4\pi^{2} 	imes 10^{-7} 	imes 200 	imes 500 	imes 10^{-4}}{0.4} 	imes (5t^{2} - 2t + 3) = (10\pi^{2} 	imes 10^{-6}) 	imes (5t^{2} - 2t + 3)$.The induced $emf$ is $\varepsilon = -\frac{d\phi}{dt} = -10\pi^{2} 	imes 10^{-6} 	imes (10t - 2)$.At $t = 1\; s$,$\varepsilon = -10\pi^{2} 	imes 10^{-6} 	imes (10(1) - 2) = -80\pi^{2} 	imes 10^{-6}\; V$.Taking magnitude,$|\varepsilon| = 80\pi^{2} 	imes 10^{-6}\; V = 80\pi^{2} 	imes 10^{-3}\; mV \approx 80 	imes 9.87 	imes 10^{-3} \approx 0.789\; mV$.Given $|\varepsilon| = \frac{4}{x} = 0.5\; mV$ (assuming $\pi^{2} \approx 10$),then $x = 8$.

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