Two long, smooth, horizontal… | JEE Advanced 2016 P1

Two long, smooth, horizontal conducting rails are 0.50 m apart and joined at one end by a resistor of 2.0 Ω. A uniform vertical magnetic field of…

One or more options correct

Two long, smooth, horizontal conducting rails are

0.50\ \mathrm{m}

apart and joined at one end by a resistor of

2.0\ \Omega

. A uniform vertical magnetic field of

0.80\ \mathrm{T}

exists in the region. A rod of mass

0.20\ \mathrm{kg}

and negligible resistance is placed across the rails and given an initial speed of

10\ \mathrm{m\,s^{-1}}

along them. The rails have negligible resistance and the rod keeps perpendicular contact. Which of the following statements is/are correct?

A. The total distance travelled by the rod before it (asymptotically) comes to rest is $25\ \mathrm{m}$ .
B. The total charge that flows through the resistor before the rod stops is $2.5\ \mathrm{C}$ .
C. The total heat generated in the resistor is $10\ \mathrm{J}$ .
D. The rod covers half of its total stopping distance in a time $mR/(B^2l^2)=2.5\ \mathrm{s}$ .

Show answer and solution

Answer: A, C

The motional emf

Blv

drives a current

I=Blv/R

, so the retarding force is

B^2l^2v/R

and

m\,dv/dt=-\dfrac{B^2l^2}{R}v

. Therefore

v=v_0e^{-t/\tau}

with

\tau=\dfrac{mR}{B^2l^2}=\dfrac{0.2\times2}{0.16}=2.5\ \mathrm{s}

. The distance is

x=v_0\tau(1-e^{-t/\tau})\to v_0\tau=25\ \mathrm{m}

t\to\infty

(A true; it takes infinite time, formally). Impulse gives

m v_0=Bl\,q

, so

q=\dfrac{mv_0}{Bl}=\dfrac{2}{0.4}=5\ \mathrm{C}

(B false:

2.5\ \mathrm{C}

is wrong). The whole initial kinetic energy turns into heat,

\tfrac12mv_0^2=10\ \mathrm{J}

(C true). In time

\tau

the rod covers

25(1-e^{-1})\approx15.8\ \mathrm{m}

, which is about

63\%

of the total, not half (D false).

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