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A coil of 180 turns carries a current of 1.5 A…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A coil of $180$ turns carries a current of $1.5$ A. The flux through each turn is $5.0 \times 10^{-4}$ Wb. Find the inductance of the coil. The current is then switched off, and it falls steadily to zero in $0.012$ s. What is the size of the emf induced in the coil?

Need a reminder? The key ideaShow

The flux linkage of a coil is $N\phi = LI$ . The self-induced emf is $\varepsilon = -L\dfrac{dI}{dt}$ , and it opposes the change of the current. For a long solenoid, $L = \dfrac{\mu_r\mu_0 N^2 A}{l}$ . The energy stored is $U = \tfrac{1}{2}LI^2$ , with $\mu_0 = 4\pi \times 10^{-7}\ \text{T m/A}$ .

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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