Question

There are three wires $MO,NO$ and $PQ$ wire. $MO$ and $NO$ are fixed and perpendicular each other, while wire $PQ$ moves with a constant velocity $v$ as shown in the figure and resistance per unit length of each wire is $\lambda$ and magnetic field exists perpendicular and inside the paper then

• A. Current in loop is induced in clockwise direction.
• B. Magnitude of current in the loop is $\frac{Bv}{\lambda (\sqrt{2}+1)}$
• C. Current in the loop is independent of time
• D. Magnitude of current decreases as time increases.

Answer 1

Answer: (B), (C)

Current in the loop is independent of time


Let $OP=OQ=s$
The coordinate of mid point of $PQ$ is
$m(\frac{x}{2},\frac{x}{2})$
$\therefore OM=\sqrt{x^2-{\left(\frac{x}{2}\right)}^2}=\frac{x}{\sqrt{2}}$
Given that velocity of $PQ$ is $v$
$\frac{d\left(OM\right)}{dt}=\frac{1}{\sqrt{2}}\frac{d\left(x\right)}{dt}=vdx=v\sqrt{2}dtx=v\sqrt{2}t\dots \left(1\right)$
Area of loop $A=\frac{1}{2}\times \text{base}\times \text{height}$
$\left(A\right)=\frac{1}{2}\times x\times x=\frac{1}{2}{x}^2=\frac{1}{2}{v}^2\times 2t^2=v^2{t}^2$ $\because \{x=\left(v\sqrt{2}\right)t\}$
Induced emf is given by
$EMF=\frac{d\Phi }{dt}=B\frac{dA}{dt}=B.\frac{d}{dt}\left(v^2{t}^2\right)=B{v}^{2}2t$
Resistance of loop = $\lambda \times \text{length}$


$R=\lambda (x+x+\sqrt{2}x)$
$R=\lambda (\sqrt{2}+2)x$
$R=\lambda (\sqrt{2}+2)v\sqrt{2}t$
Current induced in the loop is $=\frac{EMF}{R}=\frac{B{v}^{2}2t}{\lambda (2+2\sqrt{2})vt}=\frac{Bv}{\lambda (1+\sqrt{2})}$
As flux through the loop is increasing, induces current will reduce the flux as per Lenz law. Hence current induced is anticlockwise.
$b,c$ are correct