Induced emf

1. May 9, 2007

mbrmbrg

1. The problem statement, all variables and given/known data

In Figure 30-52 (see attached), a metal rod is forced to move with constant velocity v along two parallel metal rails, connected with a strip of metal at one end. A magnetic field B = 0.370 T points out of the page.

(a) If the rails are separated by 26.0 cm and the speed of the rod is 65.0 cm/s, what emf is generated?

2. Relevant equations

$$\phi_B=\oint \vec{B}\cdot d\vec{A}$$

$$\varepsilon=-\frac{d\phi_B}{dt}$$

$$v=\dot{x}$$

3. The attempt at a solution

I got the correct numerical answer, just my sign is off for part a.

$$\varepsilon=-\frac{d\phi_B}{dt}$$

$$\varepsilon=-\frac{\vec{B}\cdot d\vec{A}}{dt}$$

$$\varepsilon=-\frac{BLdx}{dt}$$

$$\varepsilon=-BLv$$

$$\varepsilon=-(0.370T)(0.26m)(0.65m/s)$$

$$\varepsilon=-0.00625V$$

Book says there is no negative in the final answer.

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2. May 9, 2007

Staff: Mentor

The minus sign in Faraday's law tells you that the direction of induced EMF opposes the change in flux. But all they want is the magnitude, which is positive. (If they wanted the direction, you'd have to specify clockwise or counterclockwise per Lenz's law.)

3. May 9, 2007

mbrmbrg

What in the problem should tip me off that they want only magnitude?

Also, what is the meaning of direction of emf? Is it the direction of the induced field lines?

4. May 9, 2007

Staff: Mentor

That they didn't ask for direction? :rofl: (Sorry!) But realize that a negative answer doesn't mean much.

You can specify the direction in terms of the direction of the induced current the EMF would drive. What would the direction of the EMF/induced current be in this problem?

Read this (especially the section on Lenz's law): Faraday's Law & Lenz's Law

5. May 9, 2007

mbrmbrg

Hmmm... I thought that unless otherwise specified, a direction was expected.

Oh. OK. I figured out that the current should be clockwise (which the problem states to be the negative direction), so the direction of the emf is clockwise/negative, as well.