In a physics book I'm reading the authors take this expression, $$\phi(t)=\displaystyle\int_0^a aB(x,t)\mathrm{d}x$$

Then they differentiate with respect to time, $$\dfrac{\mathrm{d}\phi}{\mathrm{d}t}=\displaystyle\int_0^a a\dfrac{\mathrm{d}B(x,t)}{\mathrm{d}t}\mathrm{d}x$$

I assumed that they had differentiated under the integral sign, but in that case the partial derivative of $B$ with respect to $t$ would be taken, right? But here they use the total derivative later to state that: $$\dfrac{\mathrm{d}B(x,t)}{\mathrm{d}t}=\dfrac{\partial B}{\partial t}+\dfrac{\partial B}{\partial x}\cdot\dfrac{\mathrm{d}x}{\mathrm{dt}}$$

Is this correct? If so, why?


For reference, it is a standard physics textbook for class 12 in India.


The question is:

"A square loop of side $12 \text{ cm}$ with its sides parallel to the x and y-axes is moved with a velocity of $8 \text{ cm-s}^{-1}$ in the positive x-direction in a magnetic field directed along the z-axis. This field is neither uniform in space nor constant in time. The field has a gradient of $10^{-3} \text{ T-cm}^{-1}$ and decreases with time at the rate of $10^{-3} \text{ T-s}^{-1}$. Determine the direction and magnitude of the induced current in the loop if it resistance is $4.50\text{ m}\Omega$."

  • 2
    $\begingroup$ $x$ is an integration variable. What would it mean to say that it depends on $t$? Since $x$ can't really be viewed as a function of $t$ all we can really do is treat them as independent variables and so $\frac{\mathrm{d}x}{\mathrm{d}t} = 0$ $\endgroup$ Commented Aug 13, 2017 at 14:44
  • $\begingroup$ In this case, $x$ was the the displacement along the x-axis which did vary with time because it had a velocity. $\endgroup$ Commented Aug 13, 2017 at 14:47
  • $\begingroup$ The $x$ in the expression above is can't be a function of the $t$ variable. I suggest you go back through the derivation you got this from and look very carefully at what the various quantities represent. That $t$ is not "the time at which the particle was at position $x$" (unless the writer has made a serious mistake). It may be the time at which some other even took place, or the upper limit of the integral may have a $t$ dependence (although from the expressions you have posted it doesn't seem to) the writer may also be being lazy with their variable names (this is unfortunately common) $\endgroup$ Commented Aug 13, 2017 at 14:57
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    $\begingroup$ Explicitly, which physics book is this based on, thanks? And could you edit it into your post please, as comments get deleted. $\endgroup$
    – user163104
    Commented Aug 13, 2017 at 15:03
  • $\begingroup$ @BySymmetry Could you explain why $x$ can't be a function of time? $\endgroup$ Commented Aug 13, 2017 at 16:23


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