Question

A conductor $ABQCD$ moves along its bisector with a velocity of $v=1\text{m/s}$ through a perpendicular magnetic field of $1{\text{Wb/m}}^2$, as shown in the figure. If all the four sides are of $1\text{m}$ length each, then the induced emf between points $A$ and $D$ is
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• A. $1.2\text{V}$
• B. $1.4\text{V}$
• C. $1.6\text{V}$
• D. $1.8\text{V}$

Answer 1

The correct option is B $1.4\text{V}$


No emf will be induced in the parts $AB$ and $CD$ as the length of the conductor is along the velocity.

Emf will be induced in the part $BQC$.

Induced emf between points $B$ and $C$,

$E=Bv{l}^{\prime }$

Where, $l^{\prime }=$ effective length joining the end points of the conductor.

Here, $l^{\prime }=BPC$

Now,

$BPC=\sqrt{1^2+1^2}=\sqrt{2}\text{m}$

$\therefore E=Bv{l}^{\prime }=1\times 1\times \sqrt{2}=1.4\text{V}$

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