Induced EMF in antenna

1. A magnetic dipole antenna is used to detect an electromagnetic wave. The antenna is a coil of 50 turns with radius 5.0 cm. The EM wave has frequency 870 kHz, electric field amplitude 0.50 V/m, and magnetic field amplitude 1.7 X 10-9 T.

(b) Assuming it is aligned correctly, what is the amplitude of the induced emf in the coil? (Since the wavelength of this wave is much larger than 5.0 cm, it can be assumed that at any instant the fields are uniform within the coil.)

(c) What is the amplitude of the emf induced in an electric dipole antenna of length 5.0 cm aligned with the electric field of the wave?

Homework Equations

(b) $$emf = -N\frac{\Delta \Phi}{\Delta t}$$

$$\Phi = B \times A$$

(c) ?????

The Attempt at a Solution

(b) The frequency is 870 kHz, so one oscillation takes a time of $$1.15\times 10^{-6}~s$$

$$\frac{\Delta \Phi}{\Delta t}= \frac{\Delta B}{\Delta t}~\times A = \frac{(2)(1.7\times 10^{-9}~T)}{(\frac{1}{2})(1.15\times 10^{-6}~s)}~\times \pi~(0.050~m)^2$$

Would this be the way to find $$\frac{\Delta \Phi}{\Delta t}$$

(c) I'm not sure where to start.

Last edited by a moderator:
Delta2

Homework Helper
Gold Member
For ## \frac{\Delta{\Phi}}{\Delta t}##, ## B=B_o \cos(\omega t) ##, so that ## \frac{dB}{dt}=-\omega \sin(\omega t) =\omega \cos(\omega t+\pi/2) ##, where ## \omega=2 \pi f ##. The factor ## \omega ## is the correct factor to use here. Using ## f=\frac{1}{T} ## will be missing the ## 2 \pi ## factor. ## \\ ## And (c) is simple. You just need to convert ## L=5.0## cm to meters. They give you the electric field ## E ## for the electromagnetic wave. The EMF for that case is ## \mathcal{E}=\int E \cdot dl=E L ##.

Delta2 and midgic
Thank you so much for your reply. That makes sense for part (b).

I think I can write it like this: (?)
$$B = B_0~cos(\omega~t)$$
$$\frac{dB}{dt} = -B_0~\omega~sin(\omega~t)$$
Then $$emf = -N~B_0~(2\pi~f) \times \pi~r^2$$

And for part (c)....well as you point out, that's quite straight-forward. I should have been able to do that one.

Thanks very much for your help.