Question

A conducting wire of length $l$ and mass $m$ is placed on two inclined rails as shown in figure. A current $I$ is flowing in the wire in the direction shown. When no magnetic field is present in the region, the wire is just on the verge of sliding. When a vertically upward magnetic field is switched on, the wire starts moving up the incline. The distance travelled by the wire as a function of time $t$ will be

• A. $\frac{1}{2}[\frac{IBl}{m}-2g]t^2$
• B. $\frac{1}{2}[\frac{IBl}{m}\times \frac{1}{\cos \theta }-2g\sin \theta ]t^2$
• C. $\frac{1}{2}[\frac{IBl}{m}-2g\sin \theta ]t^2$
• D. $\frac{1}{2}[\frac{IBl}{m}\times \frac{\cos 2\theta }{\cos \theta }-2g\sin \theta ]t^2$
• E. None of these

Answer 1

The correct option is D $\frac{1}{2}[\frac{IBl}{m}\times \frac{\cos 2\theta }{\cos \theta }-2g\sin \theta ]t^2$

Let the coefficient of friction between the rails and the wire be $\mu$

Initially, the wire is at the verge of sliding i.e friction force $\left(f^{\prime }\right)$ must be balancing $mgsin\theta$

$\therefore$ $\mu$ $mgcos\theta =mgsin\theta$ $⟹\mu =tan\theta$ .......(1)

Now as the current $I$ flows into the plane of paper and $B$ is in upward direction. Hence the force experienced by the wire $F=BIl$ acts in the direction shown.

Let $a$ be the acceleration of the wire.

The new friction force $f=\mu (mg$ $cos\theta +BIl$ $sin\theta )=mgsin\theta$ $+BIl\frac{si{n}^2\theta }{cos\theta }$

$\therefore$ $ma=BIl$ $cos\theta -(f+mg$ $sin\theta )$

$ma=BIl$ $cos\theta -(\frac{BIlsi{n}^2\theta }{cos\theta }+2mg$ $sin\theta )$

using $co{s}^2\theta -si{n}^2\theta =cos2\theta$

$⟹$ $a=\frac{d^{2}S}{d{t}^2}=\frac{BIl}{m}\frac{cos2\theta }{cos\theta }-2gsin\theta$

Now integrating w.r.t time, we get $\frac{dS}{dt}=$ $[\frac{BIl}{m}\frac{cos2\theta }{cos\theta }-2gsin\theta ]t$

Integrating w.r.t time again, we get distance travelled $S\left(t\right)$ $=[\frac{BIl}{m}\frac{cos2\theta }{cos\theta }-2gsin\theta ]\frac{t^2}{2}$