# How to calculate induced emf of moving rod?

I have the following exercise:

A copper rod with length L is moving on two frictionless conducting rails, in a distance a from an very long straight conductor. The rods resistance: R. Ignore any self inductance. The magnetic permeability is $$\mu_0$$

a) Find the direction and magnitude of the induced current.

I know I could probably just use faradays law finding the induced emf:
$$\mathcal{E}=-\frac{d \Phi_{B}}{d t}$$

But instead I tried using this other version, and I just can't get it working: $$\begin{array}{l}{\mathcal{E}=\oint(\overrightarrow{\boldsymbol{v}} \times \overrightarrow{\boldsymbol{B}}) \cdot d \vec{l}} \\ {\text { (all or part of a closed loop moves }} \\ {\text { in a } \overrightarrow{\boldsymbol{B}} \text { field) }}\end{array}$$ I think I'm a bit confused about the direction of the dL-vector

Here is my attempt: Let say the dl-vector is along the x-axis and we integrate from a to a+L. The B-field is into the plane, and the $$\mathcal{E}=\int_{a}^{a+L}(\vec{v} \times \vec{B}(L)) \cdot d \vec{L} =\int_{a}^{a+L}(v \hat{k} \times B(L) \hat{\jmath}) \cdot d L \hat{i}=\int_{a}^{a+L}-v \cdot B(L) d L$$ $$=\int_{a}^{a+L}-v \cdot \frac{\mu_{0} I}{2 \pi L} d L=-\frac{v \mu_{0} I}{2 \pi}(\ln (a+L)-\ln (a))$$ $$=\frac{v \mu_{0} I}{2 \pi} \ln \left(\frac{a}{a+L}\right)$$

I know this is the wrong emf, I'm supposed to get $$\mathcal{E}=\frac{v \mu_{0} I}{2 \pi} \ln \left(\frac{a+L}{L}\right)$$

And even if I chose my dL-vector to go in the other direction and integrate from a+L to a. I would get the same result.

What am I doing wrong here?

Edit: So the reason why I thought my result was wrong, was since the answer of the induced current was: $$I=\frac{v \mu_{0} I}{2 \pi R} \ln \left(\frac{a}{a+L}\right)$$

But I think it comes from faradays law, where you have a negative sign in front of the induced emf. When you want to find the current you should take the magnitude. So I shouldn't multiply with the negative sign and change the fraction in the natural log. I should leave negative sign in front.

• Yes, the minus sign on Faraday's law is due to Lenz's law (which is a manifestation of conservation of energy). When calculating the induced current, you should first obtain its magnitude, and then obtain its direction from the Lorentz force law. In this kind of calculation, the minus sign is indeed ill defined. – Bruno Anghinoni Jun 15 at 16:54
• I think, the calculation (modulo sign) is correct. But what attracts my attention is that you don't mention (apart from putting the formula) that a loop integral has to be computed. It has 4 contributions of which 3 are zero. The contributions along the rails and the loop's lower part $\perp$ to the conductor don't move, so $\vec{v}=0$, i.e. $\vec{v}\times\vec{B}=0$. Only the moving upper part remains. Finally the parametrization of the loop integral can be chosen in any direction, this choice determines the orientation of the surface which determines the direction of the generated EMF. – Frederic Thomas Jun 15 at 19:07
• @FredericThomas Thank you. But I'm not sure I follow regarding the orientation of the surface. The way we have been taught Faradays law is that you choose a direction of the area-vector perpendicular to the surface. Given the direction of that vector you can determine the direction of the induced EMF and current. Is that what you mean by the choice of orientation determines the direction of the EMF? But how do I then determine the orientation of the surface? – mhj Jun 15 at 19:35
• One can actually write $(\vec{v}\times \vec{B})\cdot d\vec{r} = B\cdot (d\vec{r}\times \vec{v})$ where $(d\vec{r}\times \vec{v})$ is a surface element with a direction perpendicular to the surface which points down or up the surface. This is what I mean with orientation of the surface. Anyway, if the right hand rule is correctly applied it is known if $(\vec{v}\times \vec{B})$ is parallel or antiparallel to $d\vec{r}$. – Frederic Thomas Jun 15 at 20:55

And even if I chose my dL-vector to go in the other direction and integrate from a+L to a. I would get the same result.

This is not correct.

Vector $$d\vec L = dL\hat i$$ is to be interpreted as $$dL$$ being the component of a step in the $$\hat i$$ direction.
As $$dL$$ is a component it can be either negative or positive.

Once you have put limits on your integration whether the step is negative or positive is fixed and you do not need to do anything more other than evaluate the integral.

As a simple example, Suppose you want to find the displacement when moving between position $$a \hat i$$ and position $$b\hat i$$ and the step is $$d\vec x = dx\hat i$$.

Going from position $$a \hat i$$ to position $$b\hat i$$ the integral that you need to evaluate is $$\displaystyle \int^{\rm b}_{\rm a} dx=(b-a)$$ giving a displacement of $$(b-a)\hat i$$.

If you went the other way from position $$b \hat i$$ to position $$a\hat i$$ the integral that you need to evaluate is $$\displaystyle\int^{\rm a}_{\rm b} dx=(a-b)$$ giving a displacement of $$(a-b)\hat i$$.

Note that in both cases the indefinite integral is $$\int dx$$ and whether you moved in the direction of $$\hat i$$ or $$-\hat i$$ was determined by the limits of integration ie the values $$a$$ and $$b$$.

In your example the integral which you need to evaluate is $$\displaystyle\int_{L_{\rm start}}^{L_{\rm finish}}-v \cdot B(L) d L$$.

Doing this when you move from $$a$$ to $$(a+L)$$ the value of the integral is $$\frac{v \mu_{0} I}{2 \pi} \ln \left(\frac{a}{a+L}\right)= - \frac{v \mu_{0} I}{2 \pi} \ln \left(\frac{a+L}{a}\right)$$ when you move from $$(a+L)$$ to $$a$$ the value of the integral is $$+ \frac{v \mu_{0} I}{2 \pi} \ln \left(\frac{a+L}{a}\right)$$

Think of the rod as a battery and the sign you would assign to the emf would differ if you move through the battery from the negative terminal to the positive terminal as compared with moving through the battery from the positive terminal to the negative terminal.

The left side of the rod is the positive terminal and the right side is the negative terminal.
Outside the rod current flows from the positive terminal to the negative terminal ie anticlockwise.

In terms of Lenz, the magnetic flux through the loop is increasing so the induced current will try and reduce that increasing flux (produced by the $$B\hat j$$ field) by flowing anticlockwise producing a magnetic field in the $$-\hat j$$ direction.