A copper connector of mass m slides down two smooth copperbars, set at an angle $α$ to the horizontal, due to gravity [See fig. (6)]. At the top of the bars are
interconnected through a resistance R. The separation between the bars is equal to l. The system is located in a uniform magnetic field of induction B, perpendicular to the plane in which the connector slides. The resistances of the bars, the connector and sliding contact as well as the self inductance of the loop are assumed to be negligible. The steady state velocity of connector is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{mgRsin α}{Bl}$

$\frac{mgRα}{B^{2} l^{2}}$

$\frac{mgRsin α}{B^{2} l^{2}}$

$\frac{mgRsin α}{{Bl}^{2}}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

The steady state will be reached when the component of gravitational force is balanced by magnetic force.
Current in the circuit $= \frac{VBl}{R}$
Magnetic force $F_{m} = iBl = \frac{{VB}^{2} l^{2}}{R}$
Gravitational force $F_{g} = mgsin α$
$\begin{array}{l} ∴ mgsin α = \frac{{vB}^{2} l^{2}}{R} \\ v = \frac{mgRsin α}{B^{2} l^{2}} \end{array}$