What is the amplitude of induced EMF in a magnetic dipole antenna?

Click For Summary

Homework Help Overview

The discussion revolves around calculating the amplitude of induced electromotive force (emf) in a magnetic dipole antenna and an electric dipole antenna, given specific parameters of an electromagnetic wave. The subject area includes electromagnetic theory and antenna physics.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the induced emf using the relationship between magnetic flux and time, with some suggesting the use of angular frequency instead of frequency alone. There is also a mention of the need to convert lengths from centimeters to meters for calculations.

Discussion Status

Some participants have provided guidance on the correct approach to finding the induced emf, particularly emphasizing the use of angular frequency. There is acknowledgment of the simplicity of part (c) of the problem, but no consensus on the complete solution for part (b) has been reached.

Contextual Notes

Participants note that the amplitude of the emf for part (b) should be reported without the negative sign, as it pertains to the amplitude rather than the direction of the induced emf.

midgic
Messages
6
Reaction score
3
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.

Thanks for your help.
 
Last edited by a moderator:
  • Like
Likes   Reactions: Delta2
Physics news on Phys.org
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 ##.
 
  • Like
Likes   Reactions: 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.
 
  • Like
Likes   Reactions: Delta2 and Charles Link
They just want the amplitude for part (b) so don't include the minus sign in the answer for the amplitude of the EMF. The minus sign is just the result of a phase shift.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 9 ·
Replies
9
Views
7K
Replies
12
Views
2K
Replies
1
Views
1K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 11 ·
Replies
11
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 7 ·
Replies
7
Views
6K