A small conducting loop of radius a and resis | vedclass

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(D) The magnetic field at a distance $x$ from the long wire is $B = \frac{\mu_0 i_0}{2 \pi x}$.Since the loop is small $(x >> a)$,the magnetic flux $\

Vedclass · Accepted Answer

none of these

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(D) The magnetic field at a distance $x$ from the long wire is $B = \frac{\mu_0 i_0}{2 \pi x}$.Since the loop is small $(x >> a)$,the magnetic flux $\phi$ through the loop is $\phi = B \cdot A = \frac{\mu_0 i_0}{2 \pi x} (\pi a^2) = \frac{\mu_0 i_0 a^2}{2x}$.The induced electromotive force $(EMF)$ $\varepsilon$ is given by Faraday's law: $\varepsilon = -\frac{d\phi}{dt} = -\frac{d}{dt} \left( \frac{\mu_0 i_0 a^2}{2x} \right) = \frac{\mu_0 i_0 a^2}{2x^2} \frac{dx}{dt} = \frac{\mu_0 i_0 a^2 v}{2x^2}$.The induced current $i$ in the loop is $i = \frac{\varepsilon}{r} = \frac{\mu_0 i_0 a^2 v}{2r x^2}$.The power dissipated in the loop is $P = i^2 r = \left( \frac{\mu_0 i_0 a^2 v}{2r x^2} \right)^2 r = \frac{(\mu_0 i_0 a^2 v)^2}{4r x^4}$.Solving for $v$: $v^2 = \frac{4 P r x^4}{(\mu_0 i_0 a^2)^2} \implies v = \frac{2 x^2}{\mu_0 i_0 a^2} \sqrt{Pr}$.Comparing this with the options,none of the given expressions match exactly because they lack the $\pi$ factor in the denominator or have incorrect coefficients. Thus,the correct choice is $(D)$.

$A$ small conducting loop of radius $a$ and resistance $r$ is pulled with velocity $v$ perpendicular to a long straight conductor carrying a current $i_0$. If a constant power $P$ is dissipated in the loop,find the variation of velocity $v$ of the loop as a function of $x$. Given that $x >> a$.

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