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 $$\vec{B}=50\vec{a_x}$$
Question:
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
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:
$$V_{emf} = 5.825mV$$
Best Answer
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?