Question

The wire $\text{PQ}$ has mass $0.5\text{kg}$ and resistance $2\Omega$, can slide on the smooth, very long horizontal parallel rails separated by a distance of $1\text{m}$. The resistance of the rails is negligible. A uniform magnetic field of magnitude $0.6\text{T}$ exists in the rectangular region, and a resistance of $3\Omega$ connects the rails outside the field region. At $t=0$, the wire $\text{PQ}$ is pushed towards the right with a speed $v=5\text{m/s}$. Find the maximum distance the wire will move.

• A. $17.5\text{m}$
• B. $24.4\text{m}$
• C. $30.5\text{m}$
• D. $34.7\text{m}$

Answer 1

The correct option is D $34.7\text{m}$

Emf induced across the wire $\text{PQ}$ is,

$E=Bvl=0.6\times v\times 1=0.6v$

$\therefore$ Current flowing in the circuit is,

$i=\frac{Bvl}{R+r}=\frac{0.6v}{5}=0.12v$

So, the force acting on the wire due to magnetic field is,

$F_B=ilB=0.12v\times 1\times 0.6=0.072v$

$\therefore$ Acceleration of the wire is,

$a=\frac{-F_B}{m}=-\frac{0.072v}{0.5}=-0.144v$

$a$ is negative because velocity of wire is decreasing.


We know that,

$a=v\frac{dv}{dx}$, where $x$ is the displacement of the wire.

$\therefore v\frac{dv}{dx}=-0.144v$

$\Rightarrow dv=-0.144dx$

Integrating both sides,

${\int }_5^{0}dv=-{\int }_0^{x}0.144dx$

$\Rightarrow [0-5]=-0.144x$

$\therefore x=34.7\text{m}$

Hence, $\left(D\right)$ is the correct answer.