Thursday, October 22, 2015

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?

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 >) .
()/()=(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.
()/()=(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.

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