A wire of length l, mass m, and resistance R slides without any friction down the parallel conducting rails of negligible resistance (figure). The rails are conneçted to each other at the bottom by a resistanceless rail parallel to the wire so that the wire and the rails form a closed rectangular conducting loop. The plane of the rails makes an angle θ with the horizontal and a uniform vertical magnetic field of induction B exists throughout the region. Find the steadystate velocity of the wire.

Question

A wire of length l, mass m, and resistance R slides without any friction down the parallel conducting rails of negligible resistance (figure). The rails are conneçted to each other at the bottom by a resistanceless rail parallel to the wire so that the wire and the rails form a closed rectangular conducting loop. The plane of the rails makes an angle θ with the horizontal and a uniform vertical magnetic field of induction B exists throughout the region. Find the steadystate velocity of the wire. (A) (m g/R) ( sin θ/B2 l2 cos 2 θ) (B) (m g/R) ( sin 2 θ/B2 l2 cos 2 θ) (C) (m g R sin θ/B2 l2 cos 2 θ) (D) m g R ( sin 2 θ/B2 l2 cos θ)

Tardigrade Admin · Accepted Answer

B I l cos θ=m g sin θ ...(i) Here induced emf across slider is (B cos θ) l v ∴ Induced current I=(B l v cos θ/R) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/460be37c23ecb17c7959c031cfcb0790-.png" /> From Eq. (i) B l cos θ (B l v cos θ/R)=m g sin θ ∴ v=(m g R sin θ/B2 l2 cos 2 θ)

$\frac{m g}{R} \frac{s i n θ}{B ^{2} l ^{2} c o s ^{2} θ}$

$\frac{m g}{R} \frac{s i n ^{2} θ}{B ^{2} l ^{2} c o s ^{2} θ}$

$\frac{m g R s i n θ}{B ^{2} l ^{2} c o s ^{2} θ}$

$m g R \frac{s i n ^{2} θ}{B ^{2} l ^{2} c o s θ}$

$B I l cos θ = m g sin θ$ ...(i)
Here induced emf across slider is $(B cos θ) l v$
$∴$ Induced current $I = \frac{Bl v c o s θ}{R}$

From Eq. (i)
$Bl cos θ \frac{Bl v c o s θ}{R} = m g sin θ$
$∴ v = \frac{m g R s i n θ}{B ^{2} l ^{2} c o s ^{2} θ}$

Rmg​B2l2cos2θsinθ​

Rmg​B2l2cos2θsin2θ​

B2l2cos2θmgRsinθ​

mgRB2l2cosθsin2θ​

BIlcosθ=mgsinθ ...(i) Here induced emf across slider is (Bcosθ)lv ∴ Induced current I=RBlvcosθ​ From Eq. (i) BlcosθRBlvcosθ​=mgsinθ ∴v=B2l2cos2θmgRsinθ​

$\frac{m g}{R} \frac{s i n θ}{B ^{2} l ^{2} c o s ^{2} θ}$

$\frac{m g}{R} \frac{s i n ^{2} θ}{B ^{2} l ^{2} c o s ^{2} θ}$

$\frac{m g R s i n θ}{B ^{2} l ^{2} c o s ^{2} θ}$

$m g R \frac{s i n ^{2} θ}{B ^{2} l ^{2} c o s θ}$

$B I l cos θ = m g sin θ$ ...(i)
Here induced emf across slider is $(B cos θ) l v$
$∴$ Induced current $I = \frac{Bl v c o s θ}{R}$

From Eq. (i)
$Bl cos θ \frac{Bl v c o s θ}{R} = m g sin θ$
$∴ v = \frac{m g R s i n θ}{B ^{2} l ^{2} c o s ^{2} θ}$