Question

Two metal strips each of length $l$ as shown are kept $b$ apart and connected to a battery of emf $ϵ$ through a resistance $R$. A wire of mass $m$ lies on it. Metal strips are smooth but floor has coeff of friction $\mu$. Find how far the wire will land after leaving the metal strips after the switch is made on in Fig.

• A. $\frac{ϵb^{2}B}{\mu Rgm}$
• B. $\frac{ϵl^{2}B}{\mu Rgm}$
• C. $\frac{ϵlbB}{\mu Rgm}$
• D. $noneofthese$

Answer 1

Answer: (C)

$\frac{ϵlbB}{\mu Rgm}$

Force on the wire $\overrightarrow{F}=IbB\hat{i}$
The frame is in xy plane.
Current i in the loop is $I=\frac{\epsilon }{R}$
$\therefore \overrightarrow{F}=\frac{\epsilon }{R}\ast bB\hat{i}$
Acceleration $a=\frac{\overrightarrow{F}}{m}=\frac{\epsilon bB}{mR}\hat{i}$
Velocity of the wire after traveling a distance on the metal strips $v^2-u^2=2aL$
$v^2-0=2\frac{\epsilon bB}{mR}L$
Once the wire lands on the floor, frictional force will act on it and the wire will decelerate.
Deceleration due to friction is $a_1=\mu g$
Distance traveled before the wire comes to rest $s=\frac{v^2}{2a_1}=\frac{2\frac{\epsilon bB}{mR}L}{2\mu g}=\frac{\epsilon bLB}{mR\mu g}$