Question

A wire cd of length $l$, mass $m$, is sliding without friction on conducting rails ax and by as shown in figure. The vertical rails are connected to one another via an external resistance $R$. The entire circuit is placed in a region of space having a uniform magnetic field $B$. The field is to the plane of circuit & directed outwards. The steady speed of rod cd is

• A. $\frac{mgR}{Bl}$
• B. $\frac{mgR}{B^2{l}^2}$
• C. $\frac{mgR}{B{l}^2}$
• D. $\frac{mgR}{B^{2}l}$

Answer 1

Answer: (B)

$\frac{mgR}{B^2{l}^2}$

At steady speed of rod, the upward force on rod CD due to magnetic field should be equal to the downward gravitational force (weight) on it.

$F_{magnetic}=F_{weight}$

$ilB=mg$ where, $i=$current in rod

$⟹i=\frac{mg}{Bl}$

Now, current in the circuit is due to magnetic induction.

EMF across CD, $E=vBl$ where $v=$steady speed of rod CD

$⟹i=\frac{E}{R}=\frac{vBl}{R}$

$⟹v=\frac{iR}{Bl}$

$⟹v=\frac{mgR}{B^2{l}^2}$

Answer-(B)