Question

If the direction of magnetic field is not varied how long will the wire move after leaving the bars?

• A. $\frac{V_{0}lB}{\mu mrg}L$
• B. $(\frac{V_{0}lB}{\mu mrg}-1)L$
• C. $(\frac{V_{0}lB}{\mu mrg}+1)L$
• D. $L$

Answer 1

Answer: (B)

$(\frac{V_{0}lB}{\mu mrg}-1)L$

Force due to magnetic bars $F_m=BIl=\frac{Bl{V}_0}{r}$
Friction force $=\mu mg$
$F_{net}=\frac{Bl{V}_0}{r}-\mu mg$
$a_{net}=\frac{Bl{V}_0}{mr}-\mu g$
Now velocity of wire after moving distance $L$ is $u^2=2a_{net}L=2(\frac{Bl{V}_0}{mr}-\mu g)L$
Let wire will move $x$ distance after leaving the bars
$v^2-u^2=2ax$
$0-2(\frac{Bl{V}_0}{mr}-\mu g)L=-2\mu gx$
$x=(\frac{Bl{V}_0}{\mu gmr}-1)L$