A conductive coil has N turns of loops with a radius r, and it has a resistance R. The coil is initially in an external homogenous magnetic field B, perpendicular to it. The direction of the magnetic field is reversed in time t.

How can I calculate the total charge going through the coil?

I imagine I first have to apply the Faraday's law of induction to get the emf ($\mathcal{E}$). This I did so that I first calculated the magnetic flux

$$\Phi_B = \int B \;dA = B \pi r^2 \left( \int\limits_0^{\pi/2} \cos(\theta) \;d\theta + \int\limits_{\pi/2}^\pi \cos(\theta) \;d\theta \right) = -2B\pi r^2$$

Then I calculated the emf using the equation

$$\mathcal{E} = -N \frac{\Delta \Phi_B}{\Delta t}$$

where simply $\Delta \Phi_B = \Phi_B$ and $\Delta t = t$.

I know that the emf is defined as the work done on a unit charge to move it once around a conductive loop, and it can be written as

$$\mathcal{E} = \frac{Fl}{q} = \frac{F2\pi r}{q}$$

but I just can't figure out how to get the total amount of charge from this. I haven't taken the resistance of the coil into account yet, so maybe that's one factor.

Can you point out what I'm missing or doing wrong?

  • $\begingroup$ Is the magnetic field perpendicular to the axis of the coil or the plane in which the loops are located? $\endgroup$ Feb 5, 2019 at 18:37
  • $\begingroup$ Perpendicular to the plane of the loops, I think. The expression is not entirely clear, but I imagine the magnetic field is going straight through the loops in the initial configuration. $\endgroup$
    – Hessu
    Feb 5, 2019 at 19:57
  • $\begingroup$ If the coil is shorted, then the general equation for current can be used $L\dot {I}+RI=emf$. $\endgroup$ Feb 5, 2019 at 21:27

1 Answer 1


You just have to write $emf = Ri$ and integrate over time since $i=dQ/dt$.

  • $\begingroup$ Thanks! Now it works. Seems like I just had a wrong way of looking at the emf definition. $\endgroup$
    – Hessu
    Feb 5, 2019 at 21:04

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