I'm having trouble figuring out a different variation of this problem. On my homework, I'm asked to find the induced EMF for the following B field: $$\vec{B}=50\vec{a_y}$$

I was able to figure out the problem with the B field being: $$\vec{B}=50\vec{a_x}$$

I want to know how does my equations change if the B field is in the \$\vec{a_y}\$ direction. I'm confused shouldn't it be the same due to symmetry :) The back of my book says I'm right for $$\vec{B}=50\vec{a_x}$$ but not for $$\vec{B}=50\vec{a_y}$$

The following is my solution to the problem if $$\vec{B}=50\vec{a_x}$$


The loop shown in Figure 9.7 is inside a uniform magnetic field \$\vec{B}=50\vec{a_x}\$ mWB/\$m^2\$. If side DC of the loop cuts the flux lines at the frequency 50Hz and the loop lies in the yz plane.

Find the following:

a)The induced EMF at \$t=1\$ms

Figure 9.7

Figure 9.7

Let's use the general form of Faraday's Law.

$$V_{emf} = -\frac{d}{dt} \int_S \!\vec{B}\cdot\vec{dS}$$

1) We can simplify the integral as $$\int_S \!\vec{B}\cdot\vec{dS}=B\cos{(\phi)}zy$$ where z and y is simply the length and width of the square conducting loop.

2) Now we have the following expression to evaluate: $$V_{emf} = -\frac{d}{dt} B\cos{(\phi)}zy$$

3) We can find \$\phi\$ by noting \$\omega=2\pi f\$ so \$\phi = \omega t\$

So... $$V_{emf} = -\frac{d}{dt} B\cos{(\omega t)}zy$$

4) Now evaluate the derivative: $$V_{emf} = -Bzy \frac{d}{dt}\cos{(\omega t)}$$ $$V_{emf} = Bzy\omega\sin{(\omega t)}$$

5) Finally plug in the known numbers to find the \$V_{emf}\$. at \$t=1ms\$

$$z=3*10^{-2}$$ $$y=4*10^{-2}$$ $$f=50Hz --> \omega = 2\pi f = 100\pi$$ $$t=1ms$$ $$B=(50*10^{-3})$$

$$V_{emf} = 100\pi(50*10^{-3})(3*10^{-2})(4*10^{-2})\sin{(100\pi (1*10^{-3}))}$$

Therefore: $$V_{emf} = 5.825mV$$

  • \$\begingroup\$ +1 - Excellent use of MathJax. Here at EE.SE, the delimiter for inline MathJax is \$ instead of the more popular $. I've replaced those for you. Please check them. I hope I didn't add any errors to the expressions. \$\endgroup\$
    – Ricardo
    Sep 17, 2014 at 1:48
  • 1
    \$\begingroup\$ Cross posted at physics.SE thus explaining the $ in place of \$ : physics.stackexchange.com/q/135859/9887 \$\endgroup\$ Sep 17, 2014 at 1:53
  • \$\begingroup\$ Alfred Centauri, I thought of posting it on both forums since it is relevant to either. I'm sorry is that a bad thing I did? I'm new to both forums. \$\endgroup\$ Sep 17, 2014 at 2:05
  • \$\begingroup\$ Ricardo, thanks! I just learned LaTeX today. Somebody at the Physics forum told me to learn it because taking a picture of my equations is to messy. haha. Thanks for the uplift. :) Thanks, I think everything is intact. Also, how do I specifically quote people so they know I responded to their comment? Is a name enough or there is someway to tag them? \$\endgroup\$ Sep 17, 2014 at 2:06
  • \$\begingroup\$ You're welcome, @TwilightSparkleTheGeek. About cross-posting: isn't very well accepted on StackExchange. Here's more info on that. A more accepted approach is to ask the question in one stack and if you don't get the answer you want, you either ask a moderator to migrate it, or delete and ask it in the other stack. \$\endgroup\$
    – Ricardo
    Sep 17, 2014 at 3:10

1 Answer 1


From the perspective of the lines of flux, if the coil is at 90 degrees to these lines, the coils area is maximized i.e. the maximum number of lines of flux flow thru the coil and the induced emf would be maximum. If the plane of the coil were rotated 90 degrees to be in line with the lines of flux, the effective area of the coil (from the perspective of the lines of flux) is zero and there will be no induced emf.

At any point in between, the "effective area" changes as a sin(angle) function, where "angle" is 90 degrees when the coil is totally across the field lines and zero when it is in line.

Does this help?

  • \$\begingroup\$ A little bit, but why sin? and not cosine? I know the two functions are related but isn't the dot product defined as a cosine between two vectors. \$\endgroup\$ Sep 17, 2014 at 14:37
  • \$\begingroup\$ It makes no difference providing you choose your angle reference accordingly. I chose 90 deg across the lines of flux to be my angle that offers maximum coil exposure to the flux hence effective area is sin(90) x actual coil area. Should I have said that reference angle was zero I'd use cos instead and get the exact same result. \$\endgroup\$
    – Andy aka
    Sep 17, 2014 at 17:00
  • \$\begingroup\$ I replaced the cosine with the sine function in the equation I got above and I got the correct answer. I don't understand why though, can you draw it out or something? Please. I turned in my assignment already but I really want to understand why the answer is that for the Vemf for B = 50a_y. :) \$\endgroup\$ Sep 18, 2014 at 5:08
  • \$\begingroup\$ You are using the same phase angle reference as me in your drawing so no need to draw again - when your angle is zero the plane of the coil has no lines of flux cutting thru it therefore the induced emf is totally zero because the projected area of the coil in the direction of flux lines is zero therefore cos(angle) cannot be correct else that would predict max emf in this position. \$\endgroup\$
    – Andy aka
    Sep 18, 2014 at 7:09

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