Physics - EMI/AC Question with Solution | TestHub

Let us refer to the coil as circuit 1 and the solenoid as circuit 2 . The field in the central region of the solenoid is uniform, so the flux through the coil is $${\Phi }_{12}=B_2{A}_1=\left({\mu }_0{n}_2{I}_2\right)A_1$$ where $n_2=N_2/l=1500$ turns $/\mathrm{m}$. The mutual inductance is $$\begin{array}{rl}& M=\frac{N_1{\Phi }_{12}}{I_2}={\mu }_0{n}_2{N}_1{A}_1\\ & =\left(4\pi \times 10^{-7}\mathrm{T}\cdot \mathrm{m}/\mathrm{A}\right)\left(1500{\mathrm{m}}^{-1}\right)(10)\left(4\times 10^{-4}{\mathrm{m}}^2\right)\\ & =7.54\times 10^{-6}\mathrm{H}\end{array}$$