Question

Two straight conducting rails form a right angle where their ends are joined. A conducting bar in contact with the rails starts at the vertex at time $t=0$ and moves with a constant velocity of $5.2\text{m/s}$ along them as shown in figure. A magnetic field with $B=0.35\text{T}$ is directed out of the page.

• A. Flux through the triangle formed by the rails and bar at $t=3\text{s}$ is $85\text{Wb}$.
• B. Emf around the triangle at $t=3\text{s}$ is $57\text{V}$.
• C. Emf around the triangle at $t=3\text{s}$ is $47\text{V}$.
• D. Flux through the triangle formed by the rails and bar at $t=3\text{s}$ is $75\text{Wb}$.

Answer 1

Answer: (A), (B)

Emf around the triangle at $t=3\text{s}$ is $57\text{V}$.

Suppose the joint of the rails is the origin of the coordinate axis.


The distance moved by the bar in time $t$,

$y=vt$

The corresponding $x$ value is,

$x=y\tan {45}^{\circ }=y$

The distance, $PQ=2x=2y=2vt$

Area of $\Delta OPQ,$

$A=\frac{1}{2}\left(2x\right)\left(y\right)=xy$

$\therefore$ Flux through the triangle,

$\varphi =BA=B\left(xy\right)$

$=B\left(vt\right)\left(vt\right)=B{v}^2{t}^2$

At $t=3\text{s},$

$\varphi =0.35\times (5.2{)}^2\times 3^2\approx 85\text{Wb}$

The magnitude of induced emf around the triangle,

$E=|-\frac{d\varphi }{dt}|=\frac{d\left(B{v}^2{t}^2\right)}{dt}$

$=B{v}^2\times 2t=2B{v}^{2}t$

At $t=3\text{s}$,

$E=2\times 0.35\times (5.20{)}^2\times 3=56.8\approx 57\text{V}$

Hence, $\left(A\right)$ and $\left(B\right)$ are the correct answers.