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(A) The mutual inductance $M$ of the solenoid and the coil is given by the formula: $M = \frac{\mu_{0} N_{1} N_{2} A}{l}$.Substituting the given value

Vedclass · Accepted Answer

$24$

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(A) The mutual inductance $M$ of the solenoid and the coil is given by the formula: $M = \frac{\mu_{0} N_{1} N_{2} A}{l}$.Substituting the given values: $N_{1} = 2000$,$N_{2} = 300$,$A = 1.2 	imes 10^{-3} \, m^2$,$l = 0.3 \, m$,and $\mu_{0} = 4\pi 	imes 10^{-7} \, T \cdot m/A$.$M = \frac{4 \pi 	imes 10^{-7} 	imes 2000 	imes 300 	imes 1.2 	imes 10^{-3}}{0.3} = 3.016 	imes 10^{-3} \, H \approx 3 	imes 10^{-3} \, H$.The induced emf $\varepsilon$ is given by $\varepsilon = -M \frac{dI}{dt}$.The current changes from $1 \, A$ to $-1 \, A$,so $dI = -1 - 1 = -2 \, A$.The time interval $dt = 0.25 \, s$.$\varepsilon = - (3 	imes 10^{-3}) 	imes \left( \frac{-2}{0.25} ight) = 3 	imes 10^{-3} 	imes 8 = 24 	imes 10^{-3} \, V = 24 \, mV$.

$A$ solenoid has $2000$ turns wound over a length of $0.3\,m$. The area of cross-section is $1.2\times10^{-3}\,m^2$. Around its central section,a coil of $300$ turns is closely wound. If an initial current of $1\,A$ is reversed in $0.25\,s$,find the emf induced in the coil in $mV$.

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