A coil having 500 turns of square shape each of side 10 cm is placed normal to a magnetic field which is increasing at 1T/s. The induced emf is:

0.1V

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0.5V

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Solution

The correct option is D 5V
Induced emf $= \frac{d ϕ}{d t}$
$ϕ = B A c o s θ$
$ϕ = B A c o s θ = B A$
$e m f = \frac{d (B A)}{d t} = \frac{d B}{d t} A$
$e m f = \frac{n A d B}{d t}$ [for 'n' = 500 turns]
$e m f = 500 \times \frac{10 \times 10}{100 \times 100} \times 1 = 5 V$

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A coil having 500 turns of square shape each of side 10 cm is placed normal to a magnetic field which is increasing at 1T/s. The induced emf is:

The correct option is D 5VInduced emf =dϕdt ϕ=BAcosθ ϕ=BAcosθ=BA emf=d(BA)dt=dBdtA emf=nAdBdt [for 'n' = 500 turns] emf=500×10×10100×100×1=5V

The correct option is D 5V
Induced emf $= \frac{d ϕ}{d t}$
$ϕ = B A c o s θ$
$ϕ = B A c o s θ = B A$
$e m f = \frac{d (B A)}{d t} = \frac{d B}{d t} A$
$e m f = \frac{n A d B}{d t}$ [for 'n' = 500 turns]
$e m f = 500 \times \frac{10 \times 10}{100 \times 100} \times 1 = 5 V$