Practice Vector Analysis - CSIR-UGC NET Offered by Unacademy

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Vector Analysis

Quick practice

Question 1 of 5

Assume a rotating electric dipole as a superposition of two oscillating dipoles, one along the x axis and the other along the y axis with the later out of phase by $90^{\circ}$ . Thus, we hus, we can write $\vec{p}=p_{0}[\cos (\omega t) \hat{x}+\sin (\omega t) \hat{y}]$ . If the total power radiated by an arbitrary source is $P=\frac{\mu_{0} \ddot{p}^{2}}{6 \pi c}$ , then the power radiated by the dipole is-

A

$\frac{\mu_{0} p_{0}^{2} \omega^{2}}{3 \pi c}$

B

$\frac{\mu_{0} p_{0}^{2} \omega^{4}}{6 \pi c}$

C

$\frac{\mu_{0} p_{0}^{2} \omega^{2}}{6 \pi c}$

D

$\frac{\mu_{0} p_{0}^{2} \omega^{4}}{3 \pi c}$