NEET 2026 — Physics solution

NEET2026Physics

A rectangular wire loop of sides 8 cm and 3 cm with a small cut, is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the plane of the loop. The emf developed across the cut, if the velocity of the loop is 2 cm s $^{- 1}$ , in a direction normal to the shorter side of the loop, will be :

A
$1.8 \times 1 0^{- 4}$ volt
B
$1.3 \times 1 0^{- 4}$ volt
C
$1.2 \times 1 0^{- 4}$ volt
D
$4.8 \times 1 0^{- 4}$ volt

Solution

When the loop moves normal to the shorter side (3 cm) with the longer side (8 cm) being the effective length cutting the field lines, the motional emf is

$e = ℓ B v = (3 \times 1 0^{- 2}) \times 0.3 \times (2 \times 1 0^{- 2})$

$= 1.8 \times 1 0^{- 4}$ V.

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