Answer to Question #234563 in Electricity and Magnetism for miel

Question

A square coil is placed in a magnetic field. The coil is composed of 450 turns of wire. As the coil is wound around, the magnetic field changes at a constant tempo from 0,2 T to 0,42 T in 6 x 10-2 seconds. The induced emf is – 15 V.

1. Calculate the change in magnetic flux in the coil.
2. Calculate the length of the side of the coil

Answer 1

Faraday's law of induction can be expressed as:

$\epsilon=-N\dfrac{d\Phi_B}{dt}=-NA\dfrac{dB}{dt}=-Nl^2\dfrac{dB}{dt}$

where $\epsilon$ is the induced emf, N is the number of turns for the coil, and since it is a squared coil we also have A=l² (length of a side to the square), and dB/dt is the rate of change for the magnetic field on the coil.

For the first par, we define and substitute for $\frac{d\Phi_B}{dt}$ :

$\dfrac{d\Phi_B}{dt}=-\dfrac{\epsilon}{N}=-(\dfrac{-15\,V}{450\,turns})=(\dfrac{1\,V}{30})=0.0\overline{3}\frac{Wb}{s}$

After that, we use again the definition for Faraday's law and we find the square length l:

$\epsilon=-Nl^2\dfrac{dB}{dt} \iff l= \sqrt{\dfrac{-\epsilon}{N\dfrac{dB}{dt}}} \\ \implies l= \sqrt{\dfrac{-(-15\frac{Wb}{s})}{(450)\dfrac{(0.42-0.20)T}{6\times10^{-2}s}}} \\ l=\sqrt{\dfrac{(15\,m^2)}{(450)\dfrac{(0.42-0.20)}{6\times10^{-2}}}}=0.0953\,m$

In conclusion, we find that the change in magnetic flux in the coil $\dfrac{d\Phi_B}{dt}=0.0\overline{3}\,\cfrac{Wb}{s}$, and the length of the side of the coil $l=0.0953\,m=9.53\,{cm}$.

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.