A plane spiral with great number N of turns wound tightly to one another is located in a uniform magnetic field perpendicular to the spiral's plane. The outside radius of the spiral's tums is equal to o. The magnetic induction varies with time as B = B₀ sin $ωt$ , where B₀ and or are constants. The amplitude of induced e.m.f. in the spiral is

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${πNa}^{2} B_{0} ω$

${πNaB}_{0}^{2} ω$

$\frac{1}{3} {πNa}^{2} B_{0} ω$

$\frac{1}{3} {πN}^{2} {aB}_{0} ω$

answer is C.

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Detailed Solution

The induced e.m,f. across dl/ no. of turns at radius r and thickness dr is
$\begin{array}{l} de = - dN \cdot \frac{dΦ}{dt} = - dN \cdot \frac{d}{dt} ({πr}^{2} B) \\ = - {dNπr}^{2} \frac{d}{dr} (B_{0} \sin ωt) \\ = - {dNπr}^{2} B_{0} ωcos ωt \end{array} But dN = \frac{N}{a} dr$
Total induced e.m.f. in the spiral
$\begin{array}{l} e = \int de = \int_{0}^{a} \frac{N}{a} dr \cdot {πωB}_{0} \cos {ωtr}^{2} \\ = \frac{1}{3} {πNB}_{0} a^{2} \cos ωt = e_{0} \cos ωt \end{array}$
Amplitude of induced e.m.f.
$= \frac{1}{3} {πNa}^{2} B_{0} ω$