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 connected 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 steady state velocity of the wire.

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$\frac{m g}{R} \frac{\sin θ}{B^{2} l^{2} \cos^{2} θ}$

$m g R \frac{\sin^{2} θ}{B^{2} l^{2} \cos θ}$

$\frac{m g}{R} \frac{\sin^{2} θ}{B^{2} l^{2} \cos^{2} θ}$

$\frac{m g R \sin θ}{B^{2} l^{2} \cos^{2} θ}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$B i l \cos θ = m g \sin θ . . . . . . (i)$

Here induced emf across slider is $e = (B \cos θ) l v$

$∴ Induced current I = \frac{B l v \cos θ}{R}$

From Eq. (i)

$B l \cos θ \frac{B l v \cos θ}{R} = m g \sin θ$

$∴ v = \frac{m g R \sin θ}{B^{2} l^{2} \cos^{2} θ}$

Watch 3-min video & get full concept clarity

No courses found

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test