Lorentz force and induced electric field

[AmanWithoutAscarf]

Homework Statement

I cannot differentiate between the torque of Lorentz force and the one caused by induced electric field, in an electrodynamics problem of a charged ring falling in non-uniform magnetic field.

Relevant Equations

$F_{B} =q\mathbf{v} \times \mathbf{B}$
$\nabla \times \mathbf{E} =-\frac{\partial \mathbf{B}}{\partial t}$
$\iint \frac{\partial \mathbf{B}}{\partial t} .dS=\oint \mathbf{E} .dl$

As far as I know, there are two causes of the rotational motion of the charged ring. T
he first torque is from Lorentz force: $$T=Q\mathbf{R\times ( v} \times \mathbf{B}_{r}) =-\frac{QR^{2}}{2}\frac{dz}{dt}\frac{d\mathbf{B}_{z}}{dz}$$ (where $\mathbf{B}_{r} =-\frac{R}{2}\frac{d\mathbf{B}_{z}}{dz}$ is the radial component of the magnetic field).
The second torque is derived from force of induced electric field: $$ F=E_{d} .Q=\frac{-d\upPhi }{dt} .\frac{1}{2\pi R} .Q=-\frac{R}{2}\frac{dB_{z}}{dt} .Q\ \Longrightarrow \ T_{1} =-\frac{QR^{2}}{2}\frac{d\mathbf{B}_{z}}{dt}$$

But it turns out that the two torque T and T1 are the same. And in the solution, only one torque was considered.

So is the Lorentz force-field the same as induced electric field?

(The picture is from https://www.numerade.com/ask/questi...nd-falls-to-the-ground-uhrough-a-non-u-67393/)

Here are all the aformentioned formulas, sorry for all language inconveniences T.T

 

[TSny]

In the reference frame of the earth, the magnetic field is static. So, there is no induced electric field in this frame. The torque on the loop is due to the Lorentz force associated with the z-component of velocity of the ring and the radial (horizontal) component of the B field.

In a reference frame falling with the ring, the ring has zero vertical velocity. So, there is no Lorentz force that acts in a direction to cause a torque. In this frame, the magnetic field is time dependent and there will be an induced electric field that produces the torque.

 

[AmanWithoutAscarf]

In the reference frame of the earth, the magnetic field is static. So, there is no induced electric field in this frame. The torque on the loop is due to the Lorentz force associated with the z-component of velocity of the ring and the radial (horizontal) component of the B field.

In a reference frame falling with the ring, the ring has zero vertical velocity. So, there is no Lorentz force that acts in a direction to cause a torque. In this frame, the magnetic field is time dependent and there will be an induced electric field that produces the torque.

Wow, I've never thought of that, but I have another confusion. In the intergral form of M-F equation, they take the whole derivative of magnetic flux over time, not the partial one. So, because B varies in position and the velocity of the ring increases over time, doesn't the derivative of magnetic flux change?

 

[TSny]

So, because B varies in position and the velocity of the ring increases over time, doesn't the derivative of magnetic flux change?
View attachment 345979

Yes, the time derivative of the magnetic flux $\Phi$ through the falling ring will generally be time-dependent.

 

[AmanWithoutAscarf]

Yes, the time derivative of the magnetic flux $\Phi$ through the falling ring will generally be time-dependent.

Oh! So in this case, the intergral form of M_F is still correct while electric field $E$ on the left side is derived from Lorentz force, gives $E = v$ x $B$. And the induced electric field will be zero $E_c = 0$ only if I consider the differential form.

I think I got it know, thank you so much!!

 

[TSny]

Oh! So in this case, the intergral form of M_F is still correct while electric field $E$ on the left side is derived from Lorentz force, gives $E = v$ x $B$. And the induced electric field will be zero $E_c = 0$ only if I consider the differential form.

I'm not sure I follow this.

From the viewpoint of the Earth frame, there is no induced electric field at any point of space since the magnetic field is time-independent in this frame. In this frame, the torque on the ring is due to the tangential component of the magnetic Lorentz force.

From the viewpoint of a frame falling with the ring, the magnetic field changes with time and there is an induced electric field. The torque on the ring in this frame is due to the electric force associated with this electric field.

As you noted, the torque is the same using either viewpoint. The torque can be written as $$\tau = -\frac{Q}{2 \pi} \dot \Phi.$$ Here, $\dot \Phi$ represents the rate of change of magnetic flux through the ring. $\dot \Phi$ may be expressed as $\pi R^2 \dot B_z$, where $\dot B_ z$ is the rate at which the z-component of magnetic field changes for an observer falling with the ring. So, $$\tau = -\frac{Q R^2}{2} \dot{B}_z.$$ Using $\tau = I \dot \omega$, we have $$\dot \omega = -\frac{Q}{2m} \dot B_z.$$ Integrate both sides with respect to time from the time of release at $z = h$ to when the ring reaches $z = 0$. This yields the final angular velocity at $z = 0$: $$\omega =- \frac{Q}{2m}[B_z(0) - B_z(h)].$$

 

[AmanWithoutAscarf]

Thanks for your explanation.
First, I understand your previous statement that there is only one source of torque in each viewpoint.

And the M-F equation in intergral form is still correct with the electric field in the left side of the equation $E=v$ x $B$, thus the torque due to $E$ and the tangential component of Lorentz force are identical.