The phase difference between the flux linkage and the induced e.m.f. in a rotating coil in a uniform magnetic field

Question

The phase difference between the flux linkage and the induced e.m.f. in a rotating coil in a uniform magnetic field (A) π (B) π / 2 (C) π / 4 (D) π / 6

Tardigrade Admin · Accepted Answer

φ=B A cos θ, where θ is the angle between area and field e=(-d φ/d t)=-B A sin θ . (d θ/d t) =B A((d θ/d t)) cos (90+θ) Here, phase difference =90°.

Q. The phase difference between the flux linkage and the induced e.m.f. in a rotating coil in a uniform magnetic field

$π$

$π /2$

$π /4$

$π /6$

$ϕ = B A cos θ$ , where $θ$ is the angle between area and field $e = \frac{- d ϕ}{d t} = - B A sin θ . \frac{d θ}{d t}$
$= B A (\frac{d θ}{d t}) cos (90 + θ)$
Here, phase difference $= 9 0^{\circ}$ .

Q. The phase difference between the flux linkage and the induced e.m.f. in a rotating coil in a uniform magnetic field

π

π/2

π/4

π/6

ϕ=BAcosθ, where θ is the angle between area and field e=dt−dϕ​=−BAsinθ.dtdθ​ =BA(dtdθ​)cos(90+θ) Here, phase difference =90∘.

$π$

$π /2$

$π /4$

$π /6$

$ϕ = B A cos θ$ , where $θ$ is the angle between area and field $e = \frac{- d ϕ}{d t} = - B A sin θ . \frac{d θ}{d t}$
$= B A (\frac{d θ}{d t}) cos (90 + θ)$
Here, phase difference $= 9 0^{\circ}$ .