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Two coils have self-inductances of 36 mH and 64 mH…

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Two coils have self-inductances of $36$ mH and $64$ mH. Their coupling coefficient is $0.50$ . Find the mutual inductance. Find the inductance when they are joined in series so that their fields aid each other, and when they oppose each other. A current of $3.0$ A flows in the aiding pair. How much energy is stored?

Need a reminder? The key ideaShow

For two coils, $N_2\phi_2 = MI_1$ and $\varepsilon_2 = -M\dfrac{dI_1}{dt}$ . A coil of $N_2$ turns and radius $r_2$ inside a long solenoid of $n$ turns in each metre has $M = \mu_0 nN_2\pi r_2^2$ , as long as the coil is no wider than the solenoid. Also $M = k\sqrt{L_1L_2}$ , and in series $L = L_1 + L_2 \pm 2M$ . The energy stored is $U = \tfrac{1}{2}LI^2$ , and each cubic metre of field holds $u = \dfrac{B^2}{2\mu_0}$ .

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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