# Time varying magnetic field, yielding two effects of induction?

$$B_s$$ is nonuniform, and it's generated from a movable source(e.g magnet or electromagnet).

A rectangular loop of area $$A$$ is stationary.

The variation of flux for this case is caused from the following:

1. Spatial movement of $$B_s$$.
2. Strength variation $$B_s$$ of due to (1).

How can I quantitatively formulate $$\varepsilon_{induced}$$ here?

I could assume that:

$$\varepsilon_{induced} =\oint \vec{E} \cdot \vec{dl} = -\frac{\delta \Phi}{\delta t}$$

to:

$$\varepsilon_{induced} =\oint \vec{E} \cdot \vec{dl} = -\int \frac{\delta B_s}{\delta t} \cdot \vec{da}$$

But how can I factor in the effects of 1&2 to $$\delta B_s$$?

From my textbook,the examples brought up had the effects(1&2) mutually exclusive, but for this case both add to $$\varepsilon_{induced}$$

• Why do you think 1 and 2 are different things? The only thing that matters is that net magnetic flux changes, whether we describe it as "magnetic field moves" or "magnetic field changes". Feb 4, 2019 at 21:05
• Because of the following: Feb 4, 2019 at 21:08
• If a magnet was stationary, and it's $B$ is nonuniform, and I decideded to move the loop, there are two effects: 1) The Lorentz force separating the charges $vBL$, and 2) The strength variation of $B$. It seems for that frame those two contribute to $\varepsilon$, why is the reference frame of the question not considering two effects as well? Ultimately they yield the same results by only considering the changes of flux in both cases(assuming all the other variables in both frames are equal). Feb 4, 2019 at 21:13
• My focus onto the effects stems from Faradays paradox' examples. I now put a microscrope on these questions: "Is the flux changing, and what causes it?." Feb 4, 2019 at 21:16
• > "why is the reference frame of the question not considering two effects as well?" Because in your question the wire does not move, so $v=0$, so the motional emf $vBL$ is zero. Feb 4, 2019 at 21:17

$$\varepsilon_{ind}=-\frac{\partial\Phi_B}{\partial t}$$
In your case, the flux is variying due to the change in two factors. Recall that $$\Phi_B=B\cdot S\cdot\cos(\alpha)$$, so the derivative would be
$$\frac{\partial\Phi_B}{\partial t}=\frac{\partial B}{\partial t}\cdot S\cdot\cos(\alpha)+B\frac{\partial S}{\partial t}\cos(\alpha)$$
• Quick question, how is the second term ($B\frac{\partial S}{\partial t}\cos(\alpha))$ different from $vBL$? We know that motional emf is zero because $v$ = $0$, I'm curious if this term represents only the spatial variation of $B$ OR considers the motional emf. Mar 24, 2019 at 19:25