91Ó°ÊÓ

Given the diagram is as follows:

The length of the bar is$L$, the mass of the bar is $m$, the separation between the rails is$d$and the applied magnetic field is$B$which is acting upward.

The direction of the current through the bar can be determined using Right-hand rule.

The magnetic field is acting upward and the bar experiences a force to the right.

If the second finger of right hand indicates the direction of the magnetic field and the middle finger indicates the direction of the force, then according to right-hand rule, the thumb indicates the velocity of the charge carriers.

Hence, in the given situation, the direction of the current will be into the page.

Given that the initial speed of the bar is$0$.

The formula to calculate the magnetic force on the bar is given by

$F=ILB.........................\left(1\right)$

Here, $F$is the magnetic force, $I$is the current through the bar, $l$is the length of the bar and $B$is the magnetic field.

The formula to calculate the acceleration of te bar is given by

$a=\frac{F}{m}....................\left(2\right)$

Substitute the expression for the force from equation (1) into equation (2) to obtain the acceleration of the bar.

localid="1650066028632" $a=\frac{ILB}{m}......................\left(3\right)$

The formula to calculate the final speed of the bar is given by

$v^2=u^2+2ad...........................\left(4\right)$

Here, $v$is the final speed, $u$is the initial speed and $d$is the distance travelled by the bar which is the length of the rails.

Substitute $0$for $u$and localid="1650066046093" $\left(\frac{ILB}{m}\right)$for $a$into equation (4) and simplify to obtain the final seed of the bar.

localid="1650066059807" $\begin{array}{rcl}{v}^2& =& 0+2\left(\frac{ILB}{m}\right)d\\ v& =& \sqrt{\frac{2ILBd}{m}}\end{array}$

The required final speed of the bar is$\sqrt{\frac{2ILBd}{m}}$.