SSC JE Electrical Engineering 22 Jan 2018 Evening Paper

Question : 128 of 200

Marks: +1, -0

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.

12.8
14.3
15.9
16.8

Validate

Solution:

Given data
d = 20cm
⇒ r = 10cm =

10/100

= 0.1 m
Rate = 20 rev/sev
e = 10 V
B = ?
Now e = ϕ × rate
10 = ϕ × 20

∴\ϕ=10/20=0.5\wb

Now magnetic field acting parallel to rotation of disc

B=\ϕ/\A=0.5/\A

Now A

= \π r^2

∴\B=0.5/{πr^2}=0.5/{3.14×(10×10^{-2})^2}

= 15.9T

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