# Faraday's Law and magnetic monopoles

The magnetic monopoles does not exist which can be shown by $\int {\vec{B} \cdot d\vec{A}} = 0$.

But in Faraday's Law of electromagnetic induction, we clearly show the EMF induced is the time rate of change of the magnetic flux, which is $E = -\frac{d\Phi_B}{dt} = -\frac{d\int{\vec{B}\cdot d\vec{A}}} {dt}$.

Now if $\int {\vec{B} \cdot d\vec{A}} = 0$ then shouldn't the induced emf be zero?

• In one case you're integrating over a closed surface, in the other over an open surface. Commented Mar 9, 2013 at 10:49
• Yep, what Michael Brown said: In a statement of Faraday's law you typically have a wire loop and integrate over the surface it bounds. In the "no monopoles statement" you integrate over a closed surface surrounding the would-be monopole. Commented Mar 9, 2013 at 10:52

This can be resolved by being clear about what surface you're integrating over. In the first equation, $$\oint {\overrightarrow{B} . \overrightarrow{dA}} = 0,$$ you're integrating over any closed surface, i.e. a surface without a hole in it, such as a sphere. The equation says that the magnetic flux coming in must equal the magnetic flux going out.
But in the second equation, $$E = \frac{d}{dt} \int_\Sigma {\overrightarrow{B} . \overrightarrow{dA}},$$ you're integrating over a surface with a hole in it, where the hole is a loop of wire, as shown in this diagram from Wikipedia: