Coil length $l=1m$, its wire length $l'=10m$, current intensity that passes in it $i = 5-2t$. This is all the given information.

I tried this:

$N$ is number of loops in the coil, $s$ is the area of the coil

According to Faraday's Law:

$\displaystyle \epsilon = -\frac{\Delta \Phi}{\Delta t}$

$\Delta \Phi = N \cdot \Delta B \cdot s$

$\displaystyle \Delta B = 4\pi \times 10^{-7}\frac{N \cdot \Delta i}{l}$

$\displaystyle N = \frac{l'}{2\pi r}$ , $s = \pi r^2$

$\displaystyle \Delta \Phi = 4\pi \times 10^{-7} \frac{(\frac{l'}{2\pi r})^{2}\cdot \pi r^2}{l} \cdot \Delta i$

$\displaystyle \Delta \Phi = 10^{-7} \frac{l'^{2}}{l}\cdot \Delta i = 10^{-5}\cdot \Delta i$

$\Delta i = 5-2\Delta t \Longrightarrow \Delta \Phi = (5-2\Delta t)\times 10^{-5}$

$\displaystyle \epsilon = -\frac{(5-2\Delta t)\times 10^{-5}}{\Delta t}$

Is that it, am I done? or did I make a mistake somewhere?


In the limit (as Δt approaches 0), the ε = dΦ/dt = (1E-5)(di/dt) = (1E-5) (2). The constant 5 does not contribute to the rate of change.

  • $\begingroup$ Thank you, I missed that. $\endgroup$ – Ammar Ibrahim1 Feb 15 at 6:49

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