Question

A coil is nested inside a larger coil. When the current in the outer coil is changed, there is an induced current in the inner coil. If we double the loop density n of the outer coil, by what factor will the induced current in the inner coil change? What about if we instead double n in the inner coil (while keeping the length constant)?

Answer 1

Answer on Question #74119, Physics / Electromagnetism

Question. A coil is nested inside a larger coil. When the current in the outer coil is changed, there is an induced current in the inner coil. If we double the loop density $n$ of the outer coil, by what factor will the induced current in the inner coil change? What about if we instead double $n_0$ in the inner coil (while keeping the length constant)?

Solution.

According to Faraday's law of induction

$$\mathcal{E}_i = -N \frac{d\Phi}{dt} = -N \frac{d}{dt} (BS) = -(n_0 l) \frac{d}{dt} (\mu_0 n IS) = -\mu_0 n S (n_0 l) \frac{dI}{dt}$$

If we double the loop density $n$ of the outer coil then

$$\mathcal{E}_{i1} = -\mu_0 2 n S (n_0 l) \frac{dI}{dt}$$ $$\frac{\mathcal{E}_{i1}}{\mathcal{E}_i} = 2 \rightarrow \mathcal{E}_{i1} = 2 \mathcal{E}_i.$$

So, the induced current in the inner coil is increased by 2 times. If we double $n_0$ in the inner coil then

$$\mathcal{E}_{i1} = -\mu_0 n S (2 n_0 l) \frac{dI}{dt}$$ $$\frac{\mathcal{E}_{i2}}{\mathcal{E}_i} = 2 \rightarrow \mathcal{E}_{i2} = 2 \mathcal{E}_i.$$

So, the induced current in the inner coil is increased by 2 times too.

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