The figure shows four wire loops, with the lengths of either $L$ or $2 L .$ All four loops will move through a region of uniform magnetic field of intensity $B$ outward from the plane of paper. The loops will move at the same constant velocity. Rank the four loops according to the increasing magnitude of the emf induced as they move through the field,

(E_{c} = E_{d}) < (E_{a} = E_{b})

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(E_{c} = E_{d}) > (E_{a} = E_{b})

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E_{c} > E_{d} > E_{b} > E_{a}

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E_{c} < E_{d} < E_{b} < E_{a}

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Solution

The correct option is B $(E_{c} = E_{d}) > (E_{a} = E_{b})$
We know that the emf induced in the loop is given by

$E = B v l$

Where, $B$ is the magnetic field

$v$ is the velocity of the loop

and $l$ is the length of the loop perpendicular to both velocity and magnetic field.

It is given in the question that magnetic field and velocities of all loops are same. Therefore, induced emf $E$ depends only upon the lengths of the loop.

It is given that

$l_{a} = l_{b} = L$

and $l_{c} = l_{d} = 2 L$

$∴$ we can say that

$E_{a} = E_{b}$

and $E_{c} = E_{d} = 2 E_{a}$

$∴ (E_{a} = E_{b}) < (E_{c} = E_{d})$

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

$Note$ :-

While writing the equation of the emf induced across a moving loop in magnetic field $(E = B v l),$ keep in mind that the side used in the equation is the side which is perpendicular to both velocity and magnetic field.

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