- #1

zenterix

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- Homework Statement
- In the figure below we have a conducting bar (green) moving through a region of uniform magnetic field ##\vec{B}=-B\hat{k}## by sliding along two frictionless conducting rails (orange) that are a distance ##w## apart and connected together by a resistor with resistance ##R##.

- Relevant Equations
- Let an external force ##\vec{F}_{ext}## be applied so that the conductor moves to the right with constant velocity ##\vec{v}=v\hat{i}##.

Below I will go through derivations I saw in some notes from MIT OCW's 8.02 "Electricity and Magnetism".

We thus have a closed loop.

Let the normal vector be ##\hat{n}=\hat{k}##.

Then, the area vector is ##\vec{A}=A\hat{k}##.

The magnetic flux through this loop is

$$\Phi_B=\vec{B}\cdot\vec{A}=(-B\hat{k})\cdot (A\hat{k})=-BA=-Bwx$$

and

$$\frac{d\Phi_B}{dt}=-Bwv=-\mathcal{E}$$

where ##v=\frac{dx}{dt}##.

Thus,

$$\mathcal{E}=Bwv >0$$

which means that relative to our choice of normal vector, the current ##I=\frac{\mathcal{E}}{R}## flows counterclockwise.

A charge carrier with charge ##q## in the moving bar experiences a magnetic force

$$\vec{F}_B=q\vec{v}\times \vec{B}$$

$$=qv\hat{i}\times B(-\hat{k})=qvB\hat{j}$$

Then there is the following snippet

At this point, let me just recap what we have done.

1) calculated magnetic flux through the loop

2) the negative of the derivative of this magnetic flux is the emf.

3) calculated the magnetic force experienced by a charge in the conducting bar.

4) verified that this force does no work

5) equated the ##x##-component of this force to the external force applied to the bar

What the heck is this induced magnetic force?

We thus have a closed loop.

Let the normal vector be ##\hat{n}=\hat{k}##.

Then, the area vector is ##\vec{A}=A\hat{k}##.

The magnetic flux through this loop is

$$\Phi_B=\vec{B}\cdot\vec{A}=(-B\hat{k})\cdot (A\hat{k})=-BA=-Bwx$$

and

$$\frac{d\Phi_B}{dt}=-Bwv=-\mathcal{E}$$

where ##v=\frac{dx}{dt}##.

Thus,

$$\mathcal{E}=Bwv >0$$

which means that relative to our choice of normal vector, the current ##I=\frac{\mathcal{E}}{R}## flows counterclockwise.

A charge carrier with charge ##q## in the moving bar experiences a magnetic force

$$\vec{F}_B=q\vec{v}\times \vec{B}$$

$$=qv\hat{i}\times B(-\hat{k})=qvB\hat{j}$$

Then there is the following snippet

The magnetic field is not actually doing the work; the external force that is pulling the bar is doing the work. In order to see why, we first observe that the emf is causing charge carriers to move upward in the bar. The charge carrier has an additional vertical component of velocity

$$\vec{v}_{carrier}=v\hat{i}+u\hat{j}$$

Therefore the magnetic force on the charge carrier is given by

$$\vec{F}_B=q\vec{v}_{carrier}\times\vec{B}=q(v\hat{i}+u\hat{j})\times B(-\hat{k})$$

$$=qvB\hat{j}-quB\hat{i}$$

The external force must exactly oppose the ##x##-component of the magnetic force in order to keep the bar moving at a constant speed.

$$\vec{F}_{ext}=quB\hat{i}$$

If a charge carrier moves from the bottom to the top of the bar in ##\Delta t## then ##w=u\Delta t##.

The bar is also moving in the positive ##x##-direction by an amount ##\Delta x=v\Delta t=\frac{vw}{u}##.

The displacement vector is ##\Delta \vec{s}=(\frac{vw}{u}\hat{i}+w\hat{j}## and we can easily show that ##\vec{F}_B\times \Delta \vec{s}=0##.

At this point, let me just recap what we have done.

1) calculated magnetic flux through the loop

2) the negative of the derivative of this magnetic flux is the emf.

3) calculated the magnetic force experienced by a charge in the conducting bar.

4) verified that this force does no work

5) equated the ##x##-component of this force to the external force applied to the bar

**And now we get to my question.**We have already determined that there is a counterclockwise current in the bar and rails.

Therefore there is an induced magnetic force ##\vec{F}_{ind}## experienced by current in the bar as the bar moves to the right given by

$$\vec{F}_{ind}=I(w\hat{j})\times (-B\hat{k})=-IwB\hat{i}=-\left ( \frac{B^2w^2v}{R}\right )\hat{i}$$

that is in the opposite direction of ##\vec{v}##.

For the bar to move at a constant velocity, the net force acting on it must be zero. This means that the external agent must supply a force

$$\vec{F}_{ext}=-\vec{F}_{ind}=\left (\frac{B^2l^2v}{R}\right )\hat{i}$$

What the heck is this induced magnetic force?

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