A conducting square loop of side L and resistance R moves in its plane with a uniform velocity v perpendicular to one of its sides. A magnetic induction B constant in time and space, pointing perpendicular and into the plane at the loop exists everywhere with half the loop outside the field, as shown in figure. The induced emf is

Electromagnetic Induction PYQ — JEE Main 2002 · Shift 0 · Q134 with Solution

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Correct answer: D

Given situation
- A square conducting loop of side $L$ and resistance $R$ is moving with uniform speed $v$ in its own plane.
- The magnetic field $\vec B$ is constant in time and space wherever it exists, and is perpendicular to the plane of the loop.
- At the shown instant, half the loop is outside the field region.
Key idea: induced emf from changing flux

The induced emf is given by Faraday’s law:
$\mathcal E = -\frac{d\Phi_B}{dt}$
where
$\Phi_B = B \times (\text{area of loop inside the field})$
since $B$ is uniform and perpendicular to the loop.
Area inside the magnetic field

As the loop moves perpendicular to one of its sides, the portion of the loop inside the field changes.

If $x$ is the width of the loop inside the field, then the area inside field is
$A = Lx$
Hence the magnetic flux is
$\Phi_B = BLx$
Rate of change of flux

Since the loop moves with speed $v$ , the width $x$ changes at the rate
$\frac{dx}{dt} = v$
in magnitude.

Therefore,
$\left|\frac{d\Phi_B}{dt}\right| = BL\frac{dx}{dt} = BLv$
So the magnitude of induced emf is
$\mathcal E = vBL$
Check options
- A: zero — incorrect, because the flux through the loop is changing as the loop enters/leaves the field region.
- B: $RvB$ — incorrect, dimensions are not of emf.
- C: $vBL/R$ — incorrect, this would correspond to current-like dependence, not emf.
- D: $vBL$ — correct.
Final answer
$\boxed{\mathcal E = vBL}$
Hence, the correct option is D.