The length of side of a square loop is 4m. This loop is placed in a uniform magnetic field of 2.5 t. Outside the loop, the magnetic field is zero and it is coming out side from magnetic field perpendicularly with velocity 2m/s. Find the value of induced emf in loop after one second.

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Solution

BV = $2.5 T$
$ϕ_{1} = 2.5 \times 4 \times 4 (i n i t i a l f l u x)$
$ϕ_{1} = \frac{25}{10} \times 16 = 40 (w e b e r)$
$ϕ = 2.5 \times 4 \times 2 w e b e r (a f t e r 1 s e c . f l u x)$
$Δ ϕ = ϕ_{2} - ϕ_{1} (C h a n g e i n f l u x)$
$Δ ϕ = 20 - 40 = - 20$
$ε = - \frac{Δ ϕ}{Δ t}; ε = - \frac{(- 20)}{1 s e c}; ε = 20 v o l t$

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The length of side of a square loop is 4m. This loop is placed in a uniform magnetic field of 2.5 t. Outside the loop, the magnetic field is zero and it is coming out side from magnetic field perpendicularly with velocity 2m/s. Find the value of induced emf in loop after one second.

BV = 2.5Tϕ1=2.5×4×4(initialflux)ϕ1=2510×16=40(weber)ϕ=2.5×4×2weber(after1sec.flux)Δϕ=ϕ2−ϕ1(Changeinflux)Δϕ=20−40=−20ε=−ΔϕΔt;ε=−(−20)1sec;ε=20volt