What is the induced emf in a radio antenna receiving an AM radio signal?

In summary, the conversation discusses the calculation of the induced emf in a radio antenna receiving an AM radio station at 800 kHz. The antenna consists of an inductor with 120 turns and a radius of 0.6cm. The induced emf is due to a magnetic field oscillating from the radio wave, which is given by \vec{B}=1.0*10^{-5}sin(2\pift)\vec{k}. The equation used to calculate the emf is emf = -Nd[flux]/dt, where d[flux] = int[B*dA]. The final answer is -0.682cos(5.03*10^{-6}t) V. There is a discussion about
  • #1
royman19
2
0

Homework Statement



A radio antenna that receives an AM radio station that emits at 800 kHz is constituted of an inductor of 120 turns with a radius of 0.6cm. The inductor is home to an iduced emf du to a magnetic field oscillating from the radio wave. The magnetic field is [tex]\vec{B}[/tex]=1.0*10[tex]^{-5}[/tex]sin(2[tex]\pi[/tex]ft)[tex]\vec{k}[/tex]. What is the induced emf in the inductor? We suppose that the magnetic field is oriented according to the inductor's axe.

Homework Equations


emf = -Nd[flux]/dt

d[flux] = int[B*dA]

The Attempt at a Solution



I tried to integrate and find the flux, but I am not that great at calculus, but then what I really don't get is that I am going to have to derive to find the emf. It seems like I am going in circles. By the way, the final answer is -0.682cos(5.03*10[tex]^{-6}[/tex]t) V.

Thanks for the help!
 
Physics news on Phys.org
  • #2
I'm checking it right now, but seems to me that B here doesn't vary with area or location or anything

I'm assuming that the d in front of [flux] is a typo, as is the 2^pi(should be 2pi)

So you're integrating B*dA...>_> if B doesn't depend on the area, it just comes out of the integral and you get B*A, which is pi*r^2

then you take the time derivative of THAT(which most certainly depends on time)times -N

Your concern would be right except you're integrating over dA, then differentiating with respect to time

Edit: This is correct, and I checked it on windows calculator so it took like 6 start overs when I clicked the wrong thing :(
 
Last edited:
  • #3
Thank you very much for the help, it works now. I was just complicating the problem. For some odd reason I was integrating the function for the magnetic field in the flux calculation, but it all works now. Thank you!
 
Last edited:

1. What is induced emf?

Induced emf, or electromotive force, is the phenomenon of creating an electric potential difference in a conductor due to a changing magnetic field. This is in accordance with Faraday's law of induction, which states that a changing magnetic field will induce an electric current in a conductor.

2. How is Faraday's law related to induced emf?

Faraday's law of induction states that the induced emf in a closed loop is equal to the negative rate of change of magnetic flux through the loop. In other words, the induced emf is directly proportional to the rate of change of the magnetic field.

3. What factors affect the magnitude of induced emf?

The magnitude of induced emf is affected by the strength of the magnetic field, the number of turns in the conductor, and the speed at which the magnetic field changes. Additionally, the orientation of the conductor with respect to the magnetic field can also impact the induced emf.

4. Can induced emf be negative?

Yes, induced emf can be negative. This occurs when the magnetic flux through the loop decreases, causing the induced emf to be in the opposite direction of the current flow. This is known as Lenz's law, which states that the direction of the induced emf will oppose the change that caused it.

5. How is induced emf used in practical applications?

Induced emf has many practical applications, including in generators, transformers, and electric motors. It is also used in technologies such as magnetic levitation and wireless charging. Additionally, Faraday's law of induction is the basis for many electromagnetic devices used in everyday life.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
303
Replies
17
Views
1K
Replies
25
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
2K
Replies
27
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
154
  • Introductory Physics Homework Help
Replies
11
Views
1K
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Advanced Physics Homework Help
Replies
2
Views
1K
Back
Top