Physics, asked by manishlodhe7073, 1 year ago

What will be the magnitude of magnetic field (in t) acting parallel to the rotation of the disc of diameter 20 cm when the magnitude of emf induced between the axis of rotation and the rim of the disc is 10 v and the angular speed of rotation of disc is 20 revolutions per second?

Answers

Answered by abhi178
4

magnitude of Induced emf = ∆Φ/∆t

here, Φ is magnetic flux .

so, Φ = B.A = B(πR²) , R is radius of disc.

given, angular frequency, ω = 20rev/s = 20 × 2π rad/s = 40πrad/s

using, ∆t = 2π/ω

so, induced emf = B(πR²)/2π/ω

= BωR²/2

given, induced emf = 10V, diameter of disc = 20cm so, R = 10cm = 0.1m

now, induced emf = BωR²/2

⇒10 = B × 40π × (0.1)²/2

⇒1 = B × 2π × 0.01

⇒B = 100/2π = 15.9 T

hence, magnitude of magnetic field is 15.9T

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