# How doI find the average induced emf in a coil given the rate of change of flux density, and the area of the coil?

The magnetic flux density changes from +10 T to -10T in 5 seconds. The area of the coil is 2.5 m^2. What is the average emf induced?

By Faraday's law, this will be equal to the change in magnetic flux linkage / the time taken for this change to occur. But the net change in magnetic flux linkage is 0, so there should be no 'net' emf induced.. Intuitively, however, I know that there will be some non-zero emf induced because the flux in the circuit is changing.

Let's call $B$ the value of the magnetic field, and let's assume that

$$B(t) = B(0) -\alpha t$$

where, here, $B(0) = 10\,T$ and $\alpha = 4\,T\cdot s^{-1}$. Then the flux of $B$ through the coil, whose area is $A = 2.5\,m^2$, is

$$\Phi(t) = A(B(0) - \alpha t)$$

Then, Faraday's law tells us that, with the appropriate orientation, this causes an electromotive force $e$ where

$$e = -\frac{\textrm{d}\Phi}{\textrm{d}t} = A\alpha$$

If you only know that $B(0) = 10\,T$ and $B(t=5\,s)=-10\,T$, then

$$\langle e \rangle = \frac{1}{T} \int\limits_0^T e\,\textrm{d} t = \frac{1}{T}\left[ -\Phi\right]_0^T = \frac{A(B(0) - B(T))}{T}$$

In both cases, the result is

$$\langle e \rangle = 10\,V$$

• But isn't flux a scalar? How can it be negative? – Saad May 10 '17 at 18:58
• Flux is a scalar, so it can be negative... – Spirine May 10 '17 at 18:59
• What does negative flux mean? How is it different from positive flux? – Saad May 10 '17 at 19:06
• The flux of a vector field represent "how much of that field passes through a giver surface". For example, considering a river, the flux of the field of speed of water represent how much water goes through a given surface. But this surface center be oriented, so that when water goes upstream (for example), flux is positive, and negative when it goes downstream. It's the same for magnetic flux. – Spirine May 10 '17 at 19:08

The net change is $-10-(10)$, that's $-20$ not zero. I guess that should clear your confusion. The rest of the answer I'm giving shows some of the steps/ideas to solve this problem.

Steps

magnetic flux density = magnetic field strength i.e. two values of $\vec{B}$ are given. Magnetic flux is defined as $\phi_B = \int \vec {B} \cdot d\vec{A}$, $\vec{A}$ is the area, whose vector points perpendicular to it's surface, "the normal".

Your question deals with a net change in $\vec{B}$ and a constant value for $\vec{A}$, and change in the flux density is one dimensional (i.e. no change in direction just magnitude), so we can drop the vector symbols and change the $\int \dots d\vec{A}$ to $A$. The dot "$\cdot$" represents scalar product between vectors by taking $\cos \theta_{AB}$ where $\theta_{AB}$ is the angle between the field and the area normal, which in your case seems to be $0$ so, $\cos \theta_{AB} = 1$

Your case simplifies to this: $$\begin {array}{l} {\Delta \phi_B} = \Delta B A \cos \theta_{AB}= \Delta B A \\ \Delta B = -20, \Delta t= 5, A = 2.5 \\ \mbox{induced emf, } \mathcal{E} =-\dfrac{ \Delta \phi_B}{\Delta t} \end {array}$$

Now you can calculate it on your own.