An AC current is driven around a loop of wire. Suppose the amplitude and frequency of the current are both doubled. By what factor does the power radiated by the antenna increase?

Expert Answers

An illustration of the letter 'A' in a speech bubbles

This is a magnetic dipole. A magnetic dipole in a vacuum radiates an average power by the following relation.

`ltPgt =(mu_0 m_0^2 omega^4)/(12 pi c^3)`

`mu_0` is the magnetic permeability of a vacuum, `m_0` is the average magnetic dipole moment, omega is the angular frequency, and `c` is the speed of light.

First we know that `m_0=Ia` , where `I` is the current and a is the area of the loop. We want to know `(< P_f >) / (< P_i >)` .

`(<P_f>)/(<P_i>)=(mu_0 I_f^2 a^2 omega_f^4)/(12 pi c^3)*(12 pi c^3)/(mu_0 I_i^2 a^2 omega_i^4)=(I_f omega_f^4)/(I_i omega_i^4)`

Plug in the changes between the final and initial configurations.

`(<P_f>)/(<P_i>)=(I_f^2 omega_f^4)/(I_i^2 omega_i^4)=((2I_i)^2 (2omega_i)^4)/(I_i^2 omega_i^4)=2^6`

The antenna radiates `2^6` times more power than initially.

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Images:
Image (1 of 1)
Approved by eNotes Editorial