# Does the magnetic force has anything to do with induced EMF in Faraday's law?

So according to faraday's law, when there is a change in magnetic flux, there is an induced EMF.

My understanding: In the reference frame of the magnet, the charge is moving, hence there is a current. Since there is a current and a magnetic field, just like the Fleming's left hand rule, there is a magnetic force which acts perpendicularly.

So, does this magnetic force have anything to do with why there is induced EMF?

There are two sorts of e-m induction, to both of which you can apply the equation $$\mathscr E =- \frac{d\Phi}{dt}.$$

(1) In the first sort, the motion of a conductor gives the charge carriers in the conductor a component of velocity at right angles to the conductor, and in a suitably directed magnetic field there will be a magnetic Lorentz force parallel to the conductor which will give rise to an emf. Specifically, in an element of conductor of directed length $$d\vec l$$, there will be a 'motional' emf given by $$d\mathscr E =- (\vec v \times \vec B).d\vec l.$$ You can show that, integrated round a closed circuit, the emf is equal to the rate of change of flux linked with the circuit, as given in the first equation above.

(2) A changing magnetic field at a point is associated with the curl of the electric field at that point by the Faraday=Maxwell equation: $$\nabla \times \vec E =-\frac {d \vec B}{dt}.$$ This integrates over a stationary closed loop to give $$\mathscr E =- \frac{\partial\Phi}{\partial t}.$$ Here, the circuit is stationary, and the emf is not due to the magnetic part of the Lorentz force.

Now to answer your specific question, "does this magnetic force have anything to do with why there is induced EMF?" It's closely related to the motional emf (that is the first sort of e-m induction above). But Fleming's LH rule is for the magnetic (motor effect) force on a conductor at right angles to the conductor. This is proportional to the velocity component of the charge carriers parallel to the conductor (and hence to the current). The induced emf is proportional to the velocity component of the carriers at right angles to the conductor, which is why the conductor has to be moving in order for there to be an induced emf. Both the motor effect and the motional emf are manifestations of the magnetic Lorentz force on moving charge carriers: $$\vec F =q \vec v \times \vec B.$$ Suppose that the emf is induced in a rod that lies in the y direction and that it is being moved in the x direction. There is a magnetic field of magnitude B in the z direction. Then the velocity of the charge carriers in the wire is $$v_x\vec i+v_y\vec j$$ in which $$v_y$$ is proportional to the current in the wire. So $$\vec F=q(v_x\vec i+v_y\vec j)\times B\vec k=-qv_xB\vec j+qv_yB\vec i$$ So $$F_y=-qv_xB\ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ F_x=qv_yB$$ $$F_x$$ is responsible for the motor effect force on the wire. $$F_y$$ is responsible for the emf induced in the wire, since $$\mathscr E =\frac 1q F_y dl= v_xB\ dl$$.

• Thank you! could you explain more about the parallel and right angle part? May 28 at 14:30
• Are you familiar with the force on a moving charged particle in a magnetic field (the magnetic Lorentz force)? And are you happy with vector product notation (as used in my answer)? May 28 at 15:05
• yea sure go ahead May 28 at 15:09
• So does that mean that using the 1st method is wrong? But even though the circuit is stationary, in the reference frame of the magnet which is moving, the circuit is moving relative to the magnetic field. Hence can't I use the 1st method also of magnetic force? May 28 at 15:20
• (a) Sorry: what's the 'first method'? (b) And can we agree on a clear set-up? May I suggest a rod in the y direction that we move with velocity $v_x$ in the x direction, with a magnetic field in the z direction. [On paper, x to the right, y upwards, z out of the paper towards us.] May 28 at 15:33