Missing minus sign - Motional EMF

[Nick89]

Homework Statement

An infinitely long conductor carrying a current I in the positive z-direction lies on the z-axis.
A conducting triangle of wires lies in the yz-plane, which moves with a speed v in the positive y-direction.
The following figure shows the situation on time t = 0:

a) Calculate the induced voltages $V_Q-V_P$ and $V_P-V_R$ on t=0.
I got this, no problem.

b) Calculate the induced voltage $V_Q-V_R$ on time t=0.
Hint: First calculate the induced voltage on a little segment dl and notice that $dl = \sqrt{dz^2 + dy^2} = \sqrt{2} dy$.

Homework Equations

$$\epsilon = \int (\vec{v} \times \vec{B}) \cdot \vec{dl}$$
$$B = \frac{ \mu_0 I}{2 \pi y}$$ (at a distance y, in the negative x-direction (into the screen)

The Attempt at a Solution

I got the answer right apart from a minus sign...

I first checked the direction of the crossproduct v*B and got it to be in the positive z-direction.
Dot product with dl is then
$$( \vec{v} \times \vec{B} ) \cdot \vec{dl} = vB \, dl \cos \theta = vB \, dl cos(135) = -\frac{1}{2} \sqrt{2} vB \, dl$$
Because dl is pointing from Q to R, the angle is 90 + 45 = 135, right?

Then:
$$\epsilon = -\frac{1}{2} \sqrt{2} \int_a^{2a} B v dl = -\frac{1}{2} \sqrt{2} \int_a^{2a} \frac{\mu_0 I}{2 \pi y} v \sqrt{2} dy = - \frac{\mu_0 I v}{2 \pi} \int_a^{2a} \frac{dy}{y} = - \frac{\mu_0 I v}{2 \pi} \ln{2}$$

As I said, the answer is correct apart from the minus sign... Where did I go wrong?The answer I am given is this exactly:
$$d \epsilon = v \frac{\mu_0 I}{2 \pi y} \frac{1}{2}\sqrt{2} dl$$
$$V_Q-V_R = \int_a^{2a}\frac{\mu_0 I v}{2 \pi y} dy = \frac{ \mu_0 I v}{2 \pi } \ln{2}$$
That's all, nothing more... I can't quite figure out what they are doing, which equations they are using...

 

[Doc Al]

Homework Equations

$$\epsilon = \int (\vec{v} \times \vec{B}) \cdot \vec{dl}$$

This is missing a minus sign. Read this: http://hyperphysics.phy-astr.gsu.edu/Hbase/pegrav.html#pen"

 

[Moetasim]

Thank you for providing the context and your attempt at a solution. I would like to point out a few things that could help you understand the solution provided and where you may have gone wrong.

Firstly, it is important to remember that the motional EMF is always in the direction that opposes the change in magnetic flux. In this case, the wire moving in the positive y-direction will create a magnetic field in the negative x-direction. This means that the induced EMF will be in the positive x-direction, opposite to the direction of the current in the wire.

In your solution, you correctly identified the direction of the cross product as being in the positive z-direction. However, the dot product with dl should be negative, as the angle between the two vectors is greater than 90 degrees. This means that the correct expression for the induced EMF should be:

\epsilon = -\frac{1}{2} \sqrt{2} \int_a^{2a} B v dl = -\frac{1}{2} \sqrt{2} \int_a^{2a} \frac{\mu_0 I}{2 \pi y} v \sqrt{2} dy = \frac{\mu_0 I v}{2 \pi} \ln{2}

Note the negative sign in front of the integral.

In the solution provided, they have calculated the induced EMF on a little segment dl using the expression:

d \epsilon = v \frac{\mu_0 I}{2 \pi y} \frac{1}{2}\sqrt{2} dl

This is the expression for the induced EMF on a small segment of the wire, which is then integrated over the length of the wire to get the total induced EMF. The reason for the factor of 1/2√2 is because the segment dl is at an angle of 45 degrees with the y-axis, and thus has a length of √2dy.

The final solution is obtained by integrating this expression over the length of the wire, from a to 2a, which gives:

V_Q-V_R = \int_a^{2a}\frac{\mu_0 I v}{2 \pi y} dy = \frac{ \mu_0 I v}{2 \pi } \ln{2}

Note that there is no negative sign in front of the integral, as the negative sign was already taken into account when calculating