Question

A conducting wire $\text{PQ}$ of mass $1.8\text{kg}$ and length $2\text{m}$ can freely slide on a pair of parallel, smooth rails placed in a magnetic field of magnitude $1\text{T}$. The rails are connected by a capacitor of capacitance $0.5\text{F}$. The electric resistance of the rails and wire is zero. A constant force $F$ acts on the wire as shown. Find the value of $F$ if the acceleration of the wire is $5{\text{m/s}}^2$.

• A. $7\text{N}$
• B. $14.5\text{N}$
• C. $19\text{N}$
• D. $23\text{N}$

Answer 1

The correct option is C $19\text{N}$


Let, at any instant, the velocity of the wire is $v$.

$\therefore$ Emf induced across the ends of the wire is,

$E=Blv=1\times 2\times v=2v$

$\therefore$ Charge on capacitor is,

$q=CE=0.5\left(2v\right)=v$

Differentiating both sides,

$\frac{dq}{dt}=\frac{dv}{dt}$

$\Rightarrow i=a$

As, $a=5{\text{m/s}}^2$ $\therefore i=5\text{A}$

Where, $i$ is the current flowing through the wire.

Now applying Newton's law on the wire,

$F-F_m=ma$

Where, $F_m$ is the force due to magnetic field on the wire which acts in leftwards direction (ontained using right-hand rule).

$\therefore F-Bil=ma[\because F_m=Bil]$

$\Rightarrow F-1\times 5\times 2=1.8\times 5$

$\therefore F=9+10=19\text{N}$

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