Question

CALC A coil has400 turns and self-inductance 7.50 mH. The current in the coil varies with time according to I =.$\mathbf{680}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{cos}\left[\mathrm{\pi d}/0.0250\right]$ (a) What is the maximum emf induced in the coil? (b) What is the maximum average flux through each turn of the coil? (c)At t = 0.0180 s , what is the magnitude of the induced emf?

Short answer

A) The maximum emf induced in the coil is $\mathbf{0}\mathbf{.}\mathbf{6406}\mathit{V}$b) The maximum average flux through each turn of the coil is role="math" localid="1664181454178" $12.75\times {10}^{-6}$c) The magnitude of the induced emf is $0.494V$

Step-by-step solution

Concept of the maximum emf induced and the maximum average flux

The maximum emf induced in the coil is given as $\mathbf{E}\mathbf{=}\mathbf{-}\mathbf{L}\frac{\mathbf{di}}{\mathbf{dt}}$and the maximum average flux through each turn of the coil is given as$\mathbf{\varphi }\mathbf{=}\frac{{\mathbf{Li}}_{\mathbf{e}}}{\mathbf{N}}$

Calculate the maximum emf induced in the coil

The coil has 400 turns with self-inductance 7.50 mH and

$\mathrm{i}=680\times {10}^{-3}\cos \left[\mathrm{\pi d}/0.0250\right]$.

Substitute the values in the equation $\mathrm{E}=-\mathrm{L}\frac{\mathrm{di}}{\mathrm{dt}}$we have,

$$\begin{gathered} \mathrm{E}=-7.50\times {10}^{-3}\frac{\mathrm{d}\left(680\times {10}^{-3}\cos \left[\mathrm{\pi d}/0.0250\right]\right)}{\mathrm{dt}} \\ =5100\times {10}^{-4}\sin \left(125.6\mathrm{t}\right)\times 125.6 \\ {\mathrm{E}}_{\max }=0.6406\mathrm{V} \end{gathered}$$

Therefore, the maximum emf induced in the coil is $0.6406\mathrm{V}$

Calculate the maximum average flux through each turn of the coil

The maximum average flux through each turn of the coil is given as $\mathrm{\varphi }=\frac{{\mathrm{Li}}_{\mathrm{e}}}{\mathrm{N}}$Substitute the values

$$\begin{gathered} \mathrm{\varphi }=\frac{7.50\times {10}^{-3}\times 680\times {10}^{-3}}{400} \\ {\mathrm{\varphi }}_{\max }=12.75\times {10}^{-6} \end{gathered}$$

Therefore, the maximum average flux through each turn of the coil is$12.75\times {10}^{-6}$

Calculate the magnitude of the induced emf

From equation $\mathrm{E}=-\mathrm{L}\frac{\mathrm{di}}{\mathrm{dt}}$the magnitude of the induced emf at $t=0.0180s$is given as

$$\begin{gathered} \mathrm{E}=-7.50\times {10}^{-3}\times 680\times {10}^{-3}\left[-\mathrm{sin\pi }\left(\frac{0.0180}{0.0250}\right)\right]\times \frac{\mathrm{\pi }}{0.0250} \\ =5100\times {10}^{-6}125.6\times \sin \left(129.6\right) \\ =0.494\mathrm{V} \end{gathered}$$

Therefore, the magnitude of the induced emf is$0.494\mathrm{V}$