A small conducting loop of radius a and resis | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Answer

(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 $x >> a$.

Similar Questions

$A$ straight line conductor of length $0.4 \ m$ is moved with a speed of $7.0 \ ms^{-1}$ perpendicular to a magnetic field of intensity $0.8 \ Wb \ m^{-2}$. The induced e.m.f. across the conductor is (in $V$)

$A$ thin semicircular conducting ring of radius $R$ is falling with its plane vertical in a horizontal magnetic induction $B$. At the position $MNQ$,the speed of the ring is $V$ and the potential difference developed across the ring is

$A$ square loop of side $a$ and resistance $R$ is moved in a region of uniform magnetic field $B$ (the loop remaining completely inside the field) with a velocity $v$ through a distance $x$. The work done is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module