0
$\begingroup$

For a current carrying closed wire, when the flux of magnetic field changes, there is an emf which is induced, given by the formula:

$\epsilon= -(d\phi/dt)$

Here, $\phi$ is the magnetic flux and $\epsilon$ is the induced emf. My doubt is, why do we have a negative sign here? In Griffith's Electrodynamics, he reasons it by saying it's negative because $dx/dt$ is negative. Is there any other reason? I don't know why I can't accept velocity being negative as the valid reason.

$\endgroup$
1

2 Answers 2

1
$\begingroup$

I'll try to illustrate using the wire loop example. As the magnetic field incident on the loop varies in time, it induces an electric field within the loop. This is from Faraday's law. This electric field creates an induced current in the loop.

The induced current also creates its own magnetic field, based on Maxwell-Ampere's law. The induced current flows such that the magnetic field it creates opposes the original change in the magnetic field. This law that the induced current opposes the original magnetic field is called Lenz' law, and it is the reason for the minus sign. It must be there to conserve energy and momentum.

Consider if this were not the case - if the minus sign were not there. If we increased the underlying magnetic field over time, we induce a current in the wire. This current creates a second magnetic field in the same direction as the original field, so the overall field strength above the loop has increased. The overall field strength would keep increasing incessantly, and this would be a violation of conservation of energy.

In the words of Griffiths, "nature abhors a change in flux".

$\endgroup$
0
$\begingroup$

It is a consequence of the sign for the conventional current, and what is convention for the direction of the magnetic field regarding the North and South poles of a magnet.

Let's have a closed horizontal circuit, where $z+$ points upward and let's place a magnet inside. While we rotate the magnet in order to put the North pole down, the magnetic field $B_z$ increases (lines go from North to South). The direction of the conventional current (positive to negative) flows clockwise in the wire, when we look from above. As the direction of the curl is positive for counterclockwise rotation, the Faraday law must have opposite signs at each side of the equation: $$\nabla \times \mathbf E = -\frac{\partial \mathbf B }{\partial t}$$

Just to compare, let's verify the case of the Ampére Law.

Let's place a wire below a compass, in the direction North-South, so that the direction of the conventional current flows South to North. That is now our $+z$ direction. The needle deflect to East, and the deflection increases when the current increases.

Conventionally, the point close to the geographic North pole that attracts the North end of the needle, is considered the Earth magnetic South Pole. So, the Earth B-field points to the North. The effect of the current is to create a transverse magnetic field directed to the East, that explains the deflection, and the wire's B-field direction is transverse and counterclockwise, if viewed from North, where the vector $\mathbf J$ (density of current) points.

So we must write the Ampere law with the same sign in both sides: $$\nabla \times \mathbf B = \mu_0 \mathbf J$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.