University:
Santa Fe College
Course:
ETS 2527C | Electromechanical Components and Mechanisms
Academic year:

2024
Views:

303

Pages:

5
Author:

Jabin F.

#Average power radiated:-

= \frac{1}{S} \int_{S} \vec{S} \cdot d\vec{a} = \int_{0}^{2π} \int_{0}^{π} \frac{\mu₀m₀²w⁴sin²θ}{32π²c³}sinθdθdϕ

= \frac{\mu₀m₀²w⁴(4π)}{32π²c³}

= \frac{\mu₀m₀²w⁴}{12πc³} ∴

_{electric} = \frac{\mu₀p₀²w⁴}{12πc³} Ques. 21 Page-96 z ↑ beam of proton like infinite charged wire. E = \frac{\mu₀I}{2πr} φ^ If line charge density = λ E = \frac{λ}{2πε₀r} φ^ I = λu = λu → λ = \frac{I}{u} E = \frac{λ}{2πε₀r} = \frac{I}{2πε₀ur} \vec{B} = \frac{\mu₀I}{2πr} \hat{ϕ} \vec{S} = \frac{\vec{E} \times \vec{B}}{\mu₀} = \frac{I²}{4π²\epsilon₀ur²} \vec{S} = \frac{I²}{4π²\epsilon₀|\vec{u}|r²} #If instead of proton beam, e- beam is taken- \vec{E} = \frac{λ}{2πε₀γ} (-\hat{γ}) = \frac{I}{2πε₀ur} \vec{B} = \frac{ωμ₀I}{2πγ} (-\hat{ϕ}) \vec{S} = \frac{I²}{4π²\epsilon₀r²u} (-\hat{r} × -\hat{ϕ}) = \frac{I²}{4π²\epsilon₀r²u} \hat{z} In this case also option c is right. as proton. scalar pot'n always depends on charge → first if q=0, v=0 electrically neufal → still it is electrically neutrally wire. bcoz ion entering and ion leaving are same.